ELASTICITY. values for the mass m and smaller values for the spring constant k lead to greater values for the period.

Transcription

1 CHAPTER 0 SIMPLE HARMONIC MOTION AND ELASTICITY ANSWERS TO FOCUS ON CONCEPTS QUESTIONS. 0. m. (c) The restoring force is given by Equation 0. as F = kx, where k is the spring constant (positive). The graph of this equation is a straight line and indicates that the restoring force has a direction that is always opposite to the direction of the displacement. Thus, when x is positive, F is negative, and vice versa. m 3. (b) According to Equations 0.4 and 0., the period T is given by T = π k. Greater values for the mass m and smaller values for the spring constant k lead to greater values for the period. 4. (d) The maximum speed in simple harmonic motion is given by Equation 0.8 ( v Aω ) max =. Thus, increases in both the amplitude A and the angular frequencyω lead to an increase in the maximum speed. 5. (e) The maximum acceleration in simple harmonic motion is given by Equation 0.0 ( a Aω ) max =. A decrease in the amplitude A decreases the maximum acceleration, but this decrease is more than offset by the increase in the angular frequency ω, which is squared in Equation m/s 7. (b) The velocity has a maximum magnitude at point A, where the object passes through the position where the spring is unstrained. The acceleration at point A is zero, because the spring is unstrained there and is not applying a force to the object. The velocity is zero at point B, where the object comes to a momentary halt and reverses the direction of its travel. The magnitude of the acceleration at point B is a maximum, because the spring is maximally stretched there and, therefore, applies a force of maximum magnitude to the object m m/s m

2 Chapter 0 Answers to Focus on Concepts Questions 477. (c) The principle of conservation of mechanical energy applies in the absence of nonconservative forces, so that KE + PE = constant. Thus, the total energy is the same at all points of the oscillation cycle. At the equilibrium position, where the spring is unstrained, the potential energy is zero, and the kinetic energy is KE max ; thus, the total energy is KE max. At the extreme ends of the cycle, where the object comes to a momentary halt, the kinetic energy is zero, and the potential energy is PE max ; thus, the total energy is also PE max. Since both KE max and PE max equal the total energy, it must be true that KE max = PE max.. (e) In simple harmonic motion the speed and, hence, KE has a maximum value as the object passes through its equilibrium position, which is position. EPE has a maximum value when the spring is maximally stretched at position 3. GPE has a maximum value when the object is at its highest point above the ground, that is, at position. 3. (a) At the instant the top block is removed, the total mechanical energy of the remaining system is all elastic potential energy and is ka (see Equation 0.3), where A is the amplitude of the previous simple harmonic motion. This total mechanical energy is conserved, because friction is absent. Therefore, the total mechanical energy of the ensuing simple harmonic motion is also ka, and the amplitude remains the same as it was k previously. The angular frequency ω is given by Equation 0. as ω =. Thus, when m the mass m attached to the spring decreases, the angular frequency increases. 4. (b) The angular frequency ω of oscillation of a simple pendulum is given by Equation 0.6 g ω =. It depends only on the magnitude g of the acceleration due to gravity and the L length L of the pendulum. It does not depend on the mass. Therefore, the pendulum with the greatest length has the smallest frequency s 6. (c) When the energy of the system is dissipated, the amplitude of the motion decreases. The motion is called damped harmonic motion. 7. (a) Resonance occurs when the frequency of the external force equals the frequency of oscillation of the object on the spring. The angular frequency of such a system is given by k Equation 0. ω =. Since the frequency of the force is doubled, the new frequency m must be k 8 ω =. The frequency of system A is, in fact, m ω = k k m = m.

3 478 SIMPLE HARMONIC MOTION AND ELASTICITY ΔL 8. (c) According to Equation 0.7 F = Y A, the force F required to stretch a piece of L 0 material is proportional to Young s modulus Y, the amount of stretch L, and the crosssectional area A of the material, but is inversely proportional to the initial length L 0 of the FL0 material. Solving this equation for the amount of stretch gives Δ L =. Thus, the greater YA the cross-sectional area, the smaller is the amount of stretch, for given values of Young s modulus, the initial length, and the stretching force. Thus, B stretches more than A, because B has the smaller cross-sectional area of solid material m

4 Chapter 0 Problems 479 CHAPTER 0 SIMPLE HARMONIC MOTION AND ELASTICITY PROBLEMS. SSM REASONING AND SOLUTION Using Equation 0., we first determine the spring constant: Applied F x 89.0 N k = = = 4660 N/m x 0.09 m Again using Equation 0., we find that the force needed to compress the spring by m is F Applied x = kx= (4660 N/m)( m) = 37 N. REASONING The weight of the block causes the spring to stretch. The amount x of stretching, according to Equation 0., depends on the magnitude F Applied of the applied force and the spring constant k. SOLUTION a. The applied force is equal to the weight of the block. The amount x that the spring stretches is equal to the length of the stretched spring minus the length of the unstretched spring. The spring constant is k FApplied 4.50 N = = = x m 0.00 m 30.0 N / m (0.) b. When a block of unknown weight is attached to the spring, it stretches it by m 0.00 m. The weight (or applied force) of the block is ( )( ) FApplied = kx = 30.0 N / m m 0.00 m = 9.00 N (0.) 3. REASONING The force required to stretch the spring is given by Equation 0. as F x Applied = kx, where k is the spring constant and x is the displacement of the stretched spring from its unstrained length. Solving for the spring constant gives k = F Applied / x. The force applied to the spring has a magnitude equal to the weight W of the board, so F = W. x Applied x

5 480 SIMPLE HARMONIC MOTION AND ELASTICITY Since the board s lower end just extends to the floor, the unstrained length x 0 of the spring, plus the length L 0 of the board, plus the displacement x of the stretched spring must equal the height h 0 of the room, or x 0 + L 0 + x = h 0. Thus, x = h 0 x 0 L 0. SOLUTION Substituting Applied x Applied x F = W and x = h 0 x 0 L 0 into Equation 0., we find F W 04 N k = = = = 650 N/m x h x L.44 m.98 m 0.30 m REASONING The restoring force of the spring and the static frictional force point in opposite directions. Since the box is in equilibrium just before it begins to move, the net force in the horizontal direction is zero at this instant. This condition, together with the expression for the restoring force (Equation 0.) and the expression for the maximum static frictional force (Equation 4.7), will allow us to determine how far the spring can be stretched without the box moving upon release. SOLUTION The drawing at the right shows the four forces that act on the box: its weight mg, the normal force F N, the restoring force F x exerted by the spring, and the maximum static frictional force f s MAX. Since the box is not moving, it is in equilibrium. Let the x axis be parallel to the table top. According to Equation 4.9a, the net force ΣF x in the x direction must be zero, ΣF x = 0. The restoring force is F x = kx (Equation 0.), where k is the spring constant and x is the displacement of the spring (assumed to be in the +x direction). The magnitude of the maximum static frictional force is f s MAX = µ s F N (Equation 4.7), where µ s is the coefficient of static friction and F N is the magnitude of the normal force. Thus, the condition for equilibrium can be written as k x + µ s F = 0 or x = µ F s N N k ΣF x We can determine F N by noting that the box is also in vertical equilibrium, so that mg + F N or F = mg N ΣF y The distance that the spring is stretched from its unstrained position is ( 0.74)( 0.80 kg)( 9.80 m /s ) µ sfn µ smg x = = = = m k k 59 N /m F x +y F N MAX fs mg +x

6 Chapter 0 Problems SSM REASONING The weight of the person causes the spring in the scale to compress. The amount x of compression, according to Equation 0., depends on the magnitude F x Applied of the applied force and the spring constant k. SOLUTION a. Since the applied force is equal to the person s weight, the spring constant is Applied F x 670 N 4 k = = = N / m x m (0.) b. When another person steps on the scale, it compresses by 0.34 cm. The weight (or applied force) that this person exerts on the scale is F ( )( ) Applied 4 x kx = = N / m m = 90 N (0.) 6. REASONING According to Equation 0., the displacement x of the spring from its F unstrained length is x =, where F is the upward force that the spring applies to the k 5.0-kg object and k is the spring constant. The magnitude of x is what we seek. The spring constant is given. To use Equation 0., however, we also need the value of F, which we can obtain by applying Newton s second law of motion Σ F = ma (Equation 4.). ΣF is the net force acting on the object, m is the object s mass, and a is the object s acceleration, which is the same as the elevator s acceleration. SOLUTION The displacement x of the spring from its unstrained length is F x = (0.) k Along with the force of the spring, the other force that acts on the object is the downwardpointing weight of the object. Thus, assuming that upward is the positive direction, we can write the net force acting on the object as Σ F = F mg. Newton s second law becomes Σ F = F mg = ma or F = mg+ ma () Substituting Equation () into Equation 0., we find that F m( g+ a) (5.0 kg)(9.80 m/s m/s ) x = = = = m k k 830 N/m

7 48 SIMPLE HARMONIC MOTION AND ELASTICITY The amount that the spring stretches is the magnitude of this result or m. 7. REASONING The block is at equilibrium as it hangs on the spring. Therefore, the downward-directed weight of the block is balanced by the upward-directed force applied to the block by the spring. The weight of the block is mg, where m is the mass and g is the acceleration due to gravity. The force exerted by the spring is given by Equation 0. (F x = kx), where k is the spring constant and x is the displacement of the spring from its unstrained length. We will apply this reasoning twice, once to the single block hanging, and again to the two blocks. Although the spring constant is unknown, we will be able to eliminate it algebraically from the resulting two equations and determine the mass of the second block. SOLUTION Let upward be the positive direction. Setting the weight mg equal to the force kx exerted by the spring in each case gives m g = kx One hanging block and m g + m g = kx Two hanging blocks Dividing the equation on the right by the equation on the left, we can eliminate the unknown spring constant k and find that mg+ mg kx m x = or + = mg kx m x It is given that x /x = 3.0. Therefore, solving for the mass of the second block reveals that x m = m = ( 0.70 kg)( 3.0 ) =.4 kg x 8. REASONING The spring with the smaller spring constant (k = 33 N/m) is less stiff than the other spring (k = 59 N/m), and, therefore, stretches farther (x > x ). The end of the rod attached to the stiffer spring is higher than the other end by a vertical distance d = x x (see the drawing). The angle θ that the rod makes with the horizontal is given by the inverse sine function: x Unstretched position θ L x d

8 Chapter 0 Problems 483 d = L θ sin (.4) where L is the length of the rod. The amount x that each spring is stretched is given by x= F k (Equation 0.). In order to use this relation, we will first need to find the amount of force F exerted on the rod by each spring. The rod is in equilibrium, so the net torque on it must be zero: Σ τ = 0 (Equation 9.). We will choose the rod s center of gravity, located at the center of the uniform rod, as the axis of rotation. The rod s weight acts at this point, and so generates no torque. This leaves only the torques due to the two spring forces, which must sum to zero. Both torques have the same lever arm about the center of gravity: l = l = Lcosθ (see the drawing at right) Both forces, therefore, must also have equal magnitudes, which can be seen as follows: Σ τ = F ( Lcosθ) F ( Lcosθ ) = 0 or F = F Together, the two vertical spring forces support the rod s weight. Because their magnitudes are equal, each force must support half the rod s weight: F = F = mg, where m is the mass of the rod, and g is the magnitude of the acceleration due to gravity. SOLUTION As noted above, each spring stretches by an amount given by x= F k (Equation 0.). Therefore, the difference d in the heights between the high and low ends of the rod is F F d = x x = k k Now, using the fact that both force magnitudes are equal to half the weight of the rod ( F F mg ) = =, we obtain Rotation axis mg mg mg d = = k k k k Thus, using Equation.4 for the angle that the rod makes with the horizontal, we find F l θ L/ L/ θ l F

9 484 SIMPLE HARMONIC MOTION AND ELASTICITY d mg = = L L k k θ sin sin ( )( ).4 kg 9.80 m/s = sin = 7.0 ( 0.75 m) 33 N/m 59 N/m o 9. SSM REASONING AND SOLUTION The force that acts on the block is given by Newton's Second law, Fx = max (Equation 4.a). Since the block has a constant acceleration, the acceleration is given by Equation.8 with v 0 = 0m/s; that is, a x = d/t, where d is the distance through which the block is pulled. Therefore, the force that acts on the block is given by md Fx = max = t The force acting on the block is the restoring force of the spring. Thus, according to Equation 0., F = kx, where k is the spring constant and x is the displacement. Solving x Equation 0. for x and using the expression above for F x, we obtain x = F md = kt (7.00 kg)(4.00 m) = (45 N/m)(0.750 s) x = k The amount that the spring stretches is 0.40 m m 0. REASONING AND SOLUTION The figure at the right shows the original situation before the spring is cut. The weight W of the object stretches the string by an amount x. Applying F x Applied = kx (Equation 0.) to this situation, (in which F x Applied = W), gives W = kx () W kx

10 Chapter 0 Problems 485 The figure at the right shows the situation after the spring is cut into two segments of equal length. Let k' represent the spring constant of each half of the spring after it is cut. Now the weight W of the object stretches each segment by an amount x'. k'x' k'x' Applying W = kx to this situation gives W W = k'x' + k'x' = k'x' () Combining Equations () and () yields kx = k'x' From Conceptual Example we know that k' = k, so that kx = (k)x' Solving for x' gives x 0.60 m x ' = = = m 4 4. SSM REASONING When the ball is whirled in a horizontal circle of radius r at speed v, the centripetal force is provided by the restoring force of the spring. From Equation 5.3, the magnitude of the centripetal force is mv /r, while the magnitude of the restoring force is kx (see Equation 0.). Thus, mv = kx () r The radius of the circle is equal to ( L 0 + ΔL), where L 0 is the unstretched length of the spring and ΔL is the amount that the spring stretches. Equation () becomes mv L 0 + ΔL = k ΔL (') If the spring were attached to the ceiling and the ball were allowed to hang straight down, motionless, the net force must be zero: mg kx = 0, where kx is the restoring force of the spring. If we let Δy be the displacement of the spring in the vertical direction, then Solving for Δy, we obtain mg = kδy

11 486 SIMPLE HARMONIC MOTION AND ELASTICITY Δy = mg k () SOLUTION According to equation (') above, the spring constant k is given by k = mv ΔL(L 0 + ΔL) Substituting this expression for k into equation () gives or Δy = mgδl(l 0 + ΔL) mv = gδl(l 0 + ΔL) v Δy = (9.80 m/s )(0.00 m)(0.00 m m) (3.00 m/s) = m. REASONING The free-body diagram shows the F max +y s = μ s F N magnitudes and directions of the forces acting on the block. The weight mg acts downward. The +x maximum force of static friction f max s acts upward F N just before the block begins to slip. The force from Applied the spring F Applied F x = kx x = kx (Equation 0.) is directed to the right. The normal force F N from the wall mg points to the left. The magnitude of the maximum force of static friction is related to the magnitude of the normal force according to max f = µ F, where µ s is the coefficient of static friction. Since the block is Equation 4.7 ( ) s s N at equilibrium just before it begins to slip, the forces in the x direction must balance and the forces in the y direction must balance. The balance of forces in the two directions will provide two equations, from which we will determine the coefficient of static friction. SOLUTION Since the forces in the x direction and in the y direction must balance, we have F = kx and mg = µ F N s N Substituting the first equation into the second equation gives (.6 kg)( 9.80 m/s ) mg mg = µ F = µ s N s( kx) or µ = = = 0.79 s kx ( 50 N/m)( m) 3. REASONING AND SOLUTION From the drawing given with the problem statement, we see that the kinetic frictional force on the bottom block (#) is given by

12 Chapter 0 Problems 487 and the maximum static frictional force on the top block (#) is f k = µ k (m + m )g () MAX s µ s f = m g () Newton s second law applied to the bottom block gives F f k kx = 0 (3) Newton s second law applied to the top block gives MAX s kx 0 a. To find the compression x, we have from Equation (4) that f = (4) x = MAX f /k = µ s s m g/k = (0.900)(5.0 kg)(9.80 m/s )/(35 N/m) = m b. Solving Equation (3) for F and then using Equation () to substitute for f k, we find that F = kx + f k = kx + µ k (m + m )g F = (35 N/m)(0.407 m) + (0.600)(45.0 kg)(9.80 m/s ) = 397 N 4. REASONING AND SOLUTION In phase of the block's motion (uniform acceleration) we find that the net force on the block is F f k = ma where the force of friction is f k = µ k mg. Therefore F = m(a + µ k g), which is just the force exerted by the spring on the block, i.e., F = kx. So we have kx = m(a + µ k g) () We can find the acceleration using a vf v m/s 0 m/s = = = 0.0 m/s t s In phase of the block's motion (constant speed) a = 0 m/s, so the force exerted by the spring is kx = mµ g () so that k kx µ k = mg Using this expression for µ k in Equation () we obtain

13 488 SIMPLE HARMONIC MOTION AND ELASTICITY a. Solving for k gives kx mg kx = m a+ g ( 5.0 kg)( 0.0 m/s ) ma k = x x = 0.00 m m = b. Substituting this value for k into Equation (), we have 3 (.00 0 N/m)( m) ( 5.0 kg)( 9.80 m/s ) N/m kx µ k = = = mg 5. SSM REASONING The frequency f of the eardrum s vibration is related to its angular frequency ω via ω = π f (Equation 0.6). The maximum speed during vibration is given by vmax = Aω (Equation 0.8). We will find the frequency f in part a from Equations 0.6 and 0.8. In part b, we will find the maximum acceleration that the eardrum undergoes with the aid of a = Aω (Equation 0.0). max SOLUTION a. Substituting ω = π f (Equation 0.6) into vmax the frequency f, we obtain v 3 max max ω ( π ) 7 = Aω (Equation 0.8) and solving for.9 0 m/s v = A = A f or f = = = 730Hz π A π m ( ) b. Substituting ω = π f (Equation 0.6) into a = Aω (Equation 0.0) yields the maximum acceleration of the vibrating eardrum: max ( ) ( ) ( ) 7 amax = Aω = A π f = m π 730 Hz = 3 m/s 6. REASONING The period of the engine is the time for it to make one revolution. This time is the total time required to make one thousand revolutions divided by 000. The frequency f of the rotational motion is given by Equation 0.5 as the reciprocal of the period. The angular frequency ω is the frequency f times π (Equation 0.6). SOLUTION a. The period of the engine, which is the time to complete one revolution, is Total time s 6 T = = = s Number of revolutions 000 3

14 Chapter 0 Problems 489 b. The frequency f of the rotational motion is related to the period T by 4 f = = =.00 0 Hz T s (0.5) c. The angular frequency ω of the rotational motion is related to the frequency f by ( ) 4 5 ω = π f = π.00 0 Hz =.6 0 rad /s (0.6) 7. REASONING The force F x that the spring exerts on the block just before it is released is equal to kx, according to Equation 0.. Here k is the spring constant and x is the displacement of the spring from its equilibrium position. Once the block has been released, it oscillates back and forth with an angular frequency given by Equation 0. as ω = k/ m, where m is the mass of the block. The maximum speed that the block attains during the oscillatory motion is v max = Aω (Equation 0.8). The magnitude of the maximum acceleration that the block attains is a max = Aω (Equation 0.0). SOLUTION a. The force F x exerted on the block by the spring is Fx ( )( ) = kx= 8.0 N /m 0.0 m = 9.84 N (0.) b. The angular frequency ω of the resulting oscillatory motion is ω = k 8.0 N /m 0.5 rad /s m = kg = (0.) c. The maximum speed v max is the product of the amplitude and the angular frequency: ( )( ) vmax = Aω = 0.0 m 0.5 rad /s =.6 m/s (0.8) d. The magnitude a max of the maximum acceleration is ( )( ) amax = Aω = 0.0 m 0.5 rad /s = 3. m /s (0.0) 8. REASONING AND SOLUTION a. Since the object oscillates between ± m, the motion s amplitude of the motion is m. b. From the graph, the period is T = 4.0 s. Therefore, according to Equation 0.4,

15 490 SIMPLE HARMONIC MOTION AND ELASTICITY ω = π T = π =.6 rad/s 4.0 s c. Equation 0. relates the angular frequency to the spring constant: ω = k/ m. Solving for k we find k = ω m = (.6 rad/s) (0.80 kg) =.0 N/m d. At t =.0 s, the graph shows that the spring has its maximum displacement. At this location, the object is momentarily at rest, so that its speed is v = 0m/s. e. The acceleration of the object at t =.0 s is a maximum, and its magnitude is a max = Aω = (0.080 m)(.6 rad/s) = 0.0 m/s 9. REASONING AND SOLUTION From Conceptual Example, we know that when the spring is cut in half, the spring constant for each half is twice as large as that of the original spring. In this case, the spring is cut into four shorter springs. Thus, each of the four shorter springs with 5 coils has a spring constant of 4 40 N/m =680 N/m. The angular frequency of simple harmonic motion is given by Equation 0.: k ω = m = 680 N/m = 6.0 rad/s 46 kg 0. REASONING The spring constant k of either spring is related to the mass m of the object attached to it and the angular frequency ω of its oscillation by ω = km(equation 0.). Squaring both sides of Equation 0. and solving for the mass m, we find that m= k ω. The masses m of the two objects are unknown but identical, so we eliminate m and obtain ω or k m= k = k = k () ω ω ω To deal with the angular frequencies, we turn to the maximum velocities. The magnitude v max of the maximum velocity of either object is given by vmax = Aω (Equation 0.8), where A is the amplitude of the object s motion. Both objects have a maximum velocity of the same magnitude, so we see that Aω v max = A ω = Aω or ω = () A

16 Chapter 0 Problems 49 SOLUTION The amplitude of the motion of the mass attached to spring is twice that of the motion of the mass attached to spring : A = A. With this substitution, Equation () becomes A ω ω = = ω (3) Substituting Equation (3) into Equation () yields k kω ω A k( ω ) = = = 4k = 4 74 N/m = 696 N/m ω ( ). SSM REASONING The frequency of vibration of the spring is related to the added mass m by Equations 0.6 and 0.: f = π k m () The spring constant can be determined from Equation 0.. SOLUTION Since the spring stretches by 0.08 m when a.8-kg object is suspended from its end, the spring constant is, according to Equation 0., Applied Fx mg (.8 kg)(9.80 m/s ) k = x = x = 0.08 m = N/m Solving Equation () for m, we find that the mass required to make the spring vibrate at 3.0 Hz is k m = 4π f =.5 03 N/m 4π = 4.3 kg (3.0 Hz). REASONING The object s maximum speed v max occurs when it passes through the position where the spring is unstrained. The next instant when its maximum acceleration a max occurs is when it stops momentarily at either x = +A or x = A, where A is the amplitude of the motion. The time t that elapses between these two instants is one fourth of the period T, so the time we seek is 4 t = T ()

17 49 SIMPLE HARMONIC MOTION AND ELASTICITY The period is given by T = πω(equation 0.4), where ω is the angular frequency of the oscillations. To determine the angular frequency, we note that both the maximum speed v max and the maximum acceleration a max depend upon the angular frequency: = Aω vmax (Equation 0.8) and a = Aω (Equation 0.0). Substituting Equation 0.8 into max Equation 0.0, we eliminate the amplitude A and obtain a direct relationship between the maximum speed v max, the maximum acceleration a max, and the angular frequency ω: a a = Aω = Aω ω = v ω ω = () max ( ) max max or vmax π SOLUTION Substituting Equation () into T = (Equation 0.4) yields ω π π v T = = a a max v max Then, substituting Equation (3) into Equation (), we obtain the time t between maximum speed and maximum acceleration: max max ( ) πv max πvmax π.5 m/s t = 0.85 s 4 = = = a max amax 6.89 m/s ( ) (3) 3. REASONING k a. The angular frequency ω (in rad/s) is given by Equation 0. ω =, where k is the m spring constant and m is the mass of the object. The frequency f (in Hz) can be obtained from the angular frequency by using Equation 0.6 (ω = πf). b. The block loses contact with the spring when the amplitude of the oscillation is sufficiently large. To understand why, consider the block at the very top of its oscillation cycle. There it is accelerating downward, with the maximum acceleration a max of simple harmonic motion. Contact is maintained with the spring, as long as the magnitude of this acceleration is less than the magnitude g of the acceleration due to gravity. If a max is greater than g, the end of the spring falls away from under the block. a max is given by Equation 0.0 ( a Aω ) max =, from which we can obtain the amplitude A when a max = g. SOLUTION a. From Equations 0.6 and 0. we have

18 Chapter 0 Problems 493 k k N/m ω = π f = or f.66 Hz m = π m = π kg = b. Using Equation 0.0 with a max = g gives g a = g = Aω or A= ω max Substituting Equation 0. ω = ( k/ m) k into this result gives m ( 9.80 m/s )( kg) g g gm A = = = = = ω k N/m m 4. REASONING AND SOLUTION The cup slips when the force of static friction is overcome. So F = ma = µ s mg, where the acceleration is the maximum value for the simple harmonic motion, i.e., so that a max = Aω = A(πf) (0.0) µ s = A(πf) /g = ( m)4π (.00 Hz) /(9.80 m/s ) = REASONING We will find the work done by the spring force from W elastic = kx 0 kx f (Equation 0.), where k is the spring constant and x 0 and x f are the initial and final compressions of the spring, measured relative to its unstrained length. The distances given in the problem are measured in millimeters, and must be converted to meters before using in Equation 0. with k = 50 N/m. SOLUTION The spring is initially compressed 5.0 mm, so we have m 3 x 0 = ( 5.0 mm ) = m 000 mm To extrude the tip of the pen requires an additional compression of 6.0 mm, so the final compression of the spring is m 3 xf = x mm = 5.0 mm mm =(.0 mm ) =.0 0 m 000 mm Therefore, from Equation 0., the work done by the spring force is

19 494 SIMPLE HARMONIC MOTION AND ELASTICITY ( ) elastic = 0 f = 0 f W kx kx k x x = 50 N/m m.0 0 m = 0.0 J 3 3 ( ) ( ) ( ) 6. REASONING The work done in stretching or compressing a spring is given directly by Equation 0. as W k( x x ) =, where k is the spring constant and x 0 and x f are, 0 f respectively, the initial and final displacements of the spring from its equilibrium position. The work is positive if the restoring force and the displacement have the same direction and negative if they have opposite directions. SOLUTION a. The work done in stretching the spring from +.00 to m is ( 0 f ) ( ) ( ) ( ) W = 46.0 N /m.00 m 3.00 m.84 0 J k x x = = b. The work done in stretching the spring from 3.00 m to +.00 m is ( 0 f ) ( ) ( ) ( ) W = 46.0 N /m 3.00 m.00 m.84 0 J k x x = = + c. The work done in stretching the spring from 3.00 to m is ( 0 f ) ( ) ( ) ( ) W k x x = = 46.0 N /m 3.00 m 3.00 m = 0 J 7. REASONING As the block falls, only two forces act on it: its weight and the elastic force of the spring. Both of these forces are conservative forces, so the falling block obeys the principle of conservation of mechanical energy. We will use this conservation principle to determine the spring constant of the spring. Once the spring constant is known, Equation 0., ω = k/ m, may be used to find the angular frequency of the block s vibrations. SOLUTION a. The conservation of mechanical energy states that the final total mechanical energy E f is equal to the initial total mechanical energy E 0, or E f = E 0 (Equation 6.9a). The expression for the total mechanical energy of an object oscillating on a spring is given by Equation 0.4. Thus, the conservation of total mechanical energy can be written as mvf + I ω f + mg h f + k y f = mv + I ω + mg h k y 0 E f E 0

20 Chapter 0 Problems 495 Before going any further, let s simplify this equation by noting which variables are zero. Since the block starts and ends at rest, v f = v 0 = 0 m/s. The block does not rotate, so its angular speed is zero, ω f = ω 0 = 0 rad/s. Initially, the spring is unstretched, so that y 0 = 0 m. Setting these terms equal to zero in the equation above gives mgh + ky = mgh f f 0 Solving this equation for the spring constant k, we have that ( h ) ( kg)( 9.80 m/s )( 0.50 m) mg h k = = = 0 f y f ( 0.50 m) 58.8 N /m b. The angular frequency ω of the block s vibrations depends on the spring constant k and the mass m of the block: k 58.8 N /m ω = = =.4 rad /s (0.) m kg 8. REASONING The elastic potential energy of an ideal spring is given by Equation 0.3 PE = ky, where k is the spring constant and y is the amount of stretch or elastic compression from the spring s unstrained length. Thus, we need to determine y for an object of mass m hanging stationary from the spring. Since the object is stationary, it has no acceleration and is at equilibrium. The upward pull of the spring balances the downwardacting weight of the object. We can use this fact to determine y. SOLUTION The magnitude of the pull of the spring is given by ky, according to Equation 0.. The weight of the object is mg. Since these two forces must balance, we have ky = mg or y = mg/k. Substituting this result into Equation 0.3 for the elastic potential energy gives PE elastic mg m g = ky = k = k k Applying this expression to the two spring/object systems, we find mg mg PE and PE k k = = Here, we have omitted the subscript elastic for convenience. Dividing the right-hand equation by the left-hand equation gives

21 496 SIMPLE HARMONIC MOTION AND ELASTICITY PE PE mg k m m mg m m ( ) ( 5.0 kg ) ( 3. kg) = = or PE = PE =.8 J = 4.4 J k 9. SSM REASONING Since air resistance is negligible, we can apply the principle of conservation of mechanical energy, which indicates that the total mechanical energy of the block and the spring is the same at the instant it comes to a momentary halt on the spring and at the instant the block is dropped. Gravitational potential energy is one part of the total mechanical energy, and Equation 6.5 indicates that it is mgh for a block of mass m located at a height h relative to an arbitrary zero level. This dependence on h will allow us to determine the height at which the block was dropped. SOLUTION The conservation of mechanical energy states that the final total mechanical energy E f is equal to the initial total mechanical energy E 0. The expression for the total mechanical energy for an object on a spring is given by Equation 0.4, so that we have mv f + Iω f + mgh f + ky f = mv + Iω + mgh ky 0 E f The block does not rotate, so the angular speeds ω f and ω 0 are zero. Since the block comes to a momentary halt on the spring and is dropped from rest, the translational speeds v f and v 0 are also zero. Because the spring is initially unstrained, the initial displacement y 0 of the spring is likewise zero. Thus, the above expression can be simplified as follows: mgh + ky = mgh f f 0 E 0 The block was dropped at a height of h 0 h f above the compressed spring. Solving the simplified energy-conservation expression for this quantity gives ( 450 N/m)( 0.05 m) kyf h 0 f h = = = m or 4.8 cm mg 0.30 kg 9.80 m/s ( )( ) 30. REASONING The forces applied to the ball by the spring and by gravity are conservative forces. Friction is nonconservative, but it is being ignored. The normal force exerted on the ball by the surface on which it slides is also conservative, but it acts perpendicular to the ball s motion and does no work on the ball. Therefore, mechanical energy is conserved.

22 Chapter 0 Problems 497 SOLUTION In applying the principle of conservation of mechanical energy, we must include kinetic energy mv, gravitational potential energy mgh, and elastic potential energy kx. Thus, we have mv B + mgh B + kx B Final total mechanical energy = mv + mgh A A + kx A Initial total mechanical energy Initially the ball is at rest at point A, so that v A = 0.00 m / s. At point B the ball is no longer against the spring, and the spring is unstrained, so that the displacement of the spring from its unstrained length is x B = 0.00 m. Substituting these two zero values into Equation () gives ( ) mvb mghb mgha kx A or mv B mg ha hb kx A + = + = + () In Equation (), it is given that ha hb = m, since point B is higher than point A. Solving Equation () for the final speed v B gives A kx vb = g( ha hb) + m ( )( ) ( 675 N/m)( m) = 9.80 m/s m + = 6.55 m / s kg 3. REASONING Assuming that friction and air resistance are negligible, we can apply the principle of conservation of mechanical energy, which indicates that the total mechanical energy of the rod (or ram) and the spring is the same at the instant it contacts the staple and at the instant the spring is released. Kinetic energy mv is one part of the total mechanical energy and depends on the mass m and the speed v of the rod. The dependence on the speed will allow us to determine the speed of the ram at the instant of contact with the staple. SOLUTION The conservation of mechanical energy states that the final total mechanical energy E f is equal to the initial total mechanical energy E 0. The expression for the total mechanical energy for a spring/mass system is given by Equation 0.4, so that we have mv f + Iω f + mgh f + ky f = mv + Iω + mgh ky 0 E f Since the ram does not rotate, the angular speeds ω f and ω 0 are zero. Since the ram is initially at rest, the initial translational speed v 0 is also zero. Thus, the above expression can be simplified as follows: mv + mgh + ky = mgh + ky E 0 f f f 0 0 ()

23 498 SIMPLE HARMONIC MOTION AND ELASTICITY In falling from its initial height of h 0 to its final height of h f, the ram falls through a distance of h 0 h f = 0.0 m, since the spring is compressed m from its unstrained length to begin with and is still compressed m when the ram makes contact with the staple. Solving the simplified energy-conservation expression for the final speed v f gives ( y ) k y v g h h m ( ) 0 f = + f 0 f ( N/m) ( m) ( m) = + ( 9.80 m/s )( 0.0 m) = 4 m/s 0.40 kg 3. REASONING Since air resistance is negligible, we can apply the principle of conservation of mechanical energy, which indicates that the total mechanical energy of the pellet/spring system is the same when the pellet comes to a momentary halt at the top of its trajectory as it is when the pellet is resting on the compressed spring. The fact that the total mechanical energy is conserved will allow us to determine the spring constant. SOLUTION The conservation of mechanical energy states that the final total mechanical energy E f is equal to the initial total mechanical energy E 0. The expression for the total mechanical energy for an object on a spring is given by Equation 0.4, so that we have mv + f Iω + mgh f f + ky f = mv + 0 Iω + mgh ky 0 E f E 0 () The pellet does not rotate, so the angular speeds ω f and ω 0 are zero. Since the pellet is at rest as it sits on the spring and since the pellet comes to a momentary halt at the top of its trajectory, the translational speeds v 0 and v f are also zero. Because the spring is unstrained when the pellet reaches its maximum height, the final displacement y f of the spring is likewise zero. Thus, Equation () simplifies to : f = mgh mgh ky Solving this simplified energy-conservation expression for the spring constant k and noting that the pellet rises to distance of hf hi = 6.0 m above its position on the compressed spring, we find that ( f h 0) ( )( )( ) y ( ) mg h.0 0 kg 9.80 m/s 6.0 m k = = = 303 N/m m 33. SSM REASONING The only force that acts on the block along the line of motion is the force due to the spring. Since the force due to the spring is a conservative force, the

24 Chapter 0 Problems 499 principle of conservation of mechanical energy applies. Initially, when the spring is unstrained, all of the mechanical energy is kinetic energy, (/ )mv 0. When the spring is fully compressed, all of the mechanical energy is in the form of elastic potential energy, (/)k x max, where x max, the maximum compression of the spring, is the amplitude A. Therefore, the statement of energy conservation can be written as mv 0 = ka This expression may be solved for the amplitude A. SOLUTION Solving for the amplitude A, we obtain A = mv 0 = (.00 0 kg)(8.00 m/s) = m k 4 N/m 34. REASONING As the climber falls, only two forces act on him: his weight and the elastic force of the nylon rope. Both of these forces are conservative forces, so the falling climber obeys the conservation of mechanical energy. We will use this conservation law to determine how much the rope is stretched when it breaks his fall and momentarily brings him to rest. SOLUTION The conservation of mechanical energy states that the final total mechanical energy E f is equal to the initial total mechanical energy E 0, or E f = E 0 (Equation 6.9a). The expression for the total mechanical energy of an object oscillating on a spring is given by Equation 0.4. Thus, the conservation of total mechanical energy can be written as mvf + I ω f + mg h f + k y f = mv + I ω + mg h k y 0 E f Before going any further, let s simplify this equation by noting which variables are zero. Since the climber starts and ends at rest, v f = v 0 = 0 m/s. The climber does not rotate, so his angular speed is zero, ω f = ω 0 = 0 rad/s. Initially, the spring is unstretched, so that y 0 = 0 m. Setting these terms to zero in the equation above gives mgh + ky = mgh f f 0 Rearranging the terms in this equation, we have ( ) ky f + mg hf h = 0 0 y f m Note that the climber falls a distance of m before the rope starts to stretch, and y f is the displacement of the stretched rope. Since y f points downward, it is considered to be E 0

25 500 SIMPLE HARMONIC MOTION AND ELASTICITY negative. Thus, y f m is the total downward displacement of the falling climber, which is also equal to h f h 0. With this substitution, the equation above becomes f f ( ) ky + mgy mg m = 0 This is a quadratic equation in the variable y f. Using the quadratic formula gives y y f ( ) 4 ( ) ( m) ( k ) mg ± mg k mg = f = ( 86.0 kg)( 9.80 m /s ) ± N /m 3 ( 86.0 kg)( 9.80 m /s ) 4( )(.0 0 N /m ) ( 86.0 kg)( 9.80 m /s )( m) N /m There are two answers, y f = m and.95 m. Since the rope is stretched in the downward direction, which we have taken to be the negative direction, the displacement is.95 m. Thus, the amount that the rope is stretched is.95 m. 35. REASONING Since the surface is frictionless, we can apply the principle of conservation of mechanical energy, which indicates that the final total mechanical energy E f of the object and spring is equal to their initial total mechanical energy E 0 ; E f = E 0. This conservation equation, and the fact that the angular frequency ω of the oscillation is related to the spring constant k and mass m by ω = k/ m (Equation 0.), will permit us to find the speed of the object at the instant when the spring is stretched by m. SOLUTION Equating the total mechanical energy of the system at the instant the spring is stretched by x f = m to the total mechanical energy when the spring is compressed by x 0 = m, we have mv + f Iω + mgh f f + kx f = mv + 0 Iω + mgh kx 0 E f The object does not rotate, so the angular speeds ω f and ω 0 are zero. Since the object is initially at rest, v 0 = 0 m/s. Finally, we note that the height of the object does not change during the motion, so h f = h 0. Thus, Equation () simplifies to mvf + kx f = kx 0 Solving this expression for the final speed gives E 0 ()

26 Chapter 0 Problems 50 v k x x m f = 0 f We now recognize that the term k/ m is the angular frequency ω of the motion (see Equation 0.). With this substitution, the final speed becomes k v ( ) ( ) ( ) f = x0 xf = ω x0 xf =.3 rad/s m m = 0.50 m/s m 36. REASONING a. The acceleration of the box is zero when the net force acting on it is zero, in accord with Newton s second law of motion. The net force includes the box s weight (directed downward) and the restoring force of the spring (directed upward). The condition that the net force is zero will allow us to determine the magnitude of the spring s displacement. b. The speed of the box is zero when the spring is fully compressed, but the acceleration of the box is not zero at this instant. If the acceleration were zero, the box would be in equilibrium, and the net force on it would be zero. However, the box accelerates upward because the spring is exerting an upward force that is greater than the downward force due to the weight of the box. Thus, we cannot proceed as in part (a), so instead we will use energy conservation to determine the magnitude of the spring s displacement. SOLUTION a. The drawing at the right shows the two forces acting on the box: its weight mg and the restoring force F y exerted by the spring. At the instant the acceleration of +y F y the box is zero, it is in equilibrium. According to Equation 4.9b, the net force ΣF y in the y direction must be zero, ΣF y = 0. The restoring force is given by Equation 0. as F y = ky, where k is the spring constant and y is the displacement of the spring (assumed to be in the downward, or negative, direction). Thus, the condition for equilibrium can be written as mg mg ky + mg = 0 or y = k ΣF y Solving for the magnitude of the spring s displacement gives (.5 kg)( 9.80 m /s ) Magnitude of spring's mg = = = m displacement k 450 N /m b. The conservation of mechanical energy states that the final total mechanical energy E f is equal to the initial total mechanical energy E 0, or E f = E 0 (Equation 6.9a). The expression

27 50 SIMPLE HARMONIC MOTION AND ELASTICITY for the total mechanical energy of an object is given by Equation 0.4. Thus, the conservation of total mechanical energy can be written as mvf + I ω f + mg h f + k y f = mv + I ω + mg h k y 0 E f We can simplify this equation by noting which variables are zero. Since the box comes to a momentary halt, v f = 0 m/s. The box does not rotate, so its angular speed is zero, ω f = ω 0 = 0 rad/s. Initially, the spring is unstretched, so that y 0 = 0 m. Setting these terms equal to zero in the equation above gives f + f = mgh k y mv mgh E 0 The vertical displacement h f h 0 through which the box falls is equal to the displacement y f of the spring, so y f = h f h 0. Note that y f is negative, because h f is less than h 0. The downward-moving box compresses the spring in the downward direction, which, as usual, we take to be the negative direction. Substituting this expression for y f into the equation above and rearranging terms, we find that ( ) mv mg h 0 f h 0 k y = 0 f y f or kyf mgy f mv 0 + = 0 This is a quadratic equation in the variable y f. The solution is y ( ) ( )( ) mg ± mg 4 k mv 0 mg mg m f = = ± v + 0 ( k ) Substituting in the numbers, we find that k k k y f.5 kg 9.80 m/s.5 kg 9.80 m/s.5 kg = ± N/m 450 N/m 450 N/m ( )( ) ( )( ) ( 0.49 m/s) =. 0 m or m The positive answer is discarded because the spring is compressed downward by the falling box, so the displacement of the spring is negative. Therefore, the magnitude of the spring s displacement is m.

28 Chapter 0 Problems 503 k 37. SSM REASONING The angular frequency ω (in rad/s) is given by ω = m (Equation 0.), where k is the spring constant and m is the mass of the object. However, we are given neither k nor m. Instead, we are given information about how much the spring is compressed and the launch speed of the object. Once launched, the object has kinetic energy, which is related to its speed. Before launching, the spring/object system has elastic potential energy, which is related to the amount by which the spring is compressed. This suggests that we apply the principle of conservation of mechanical energy in order to use the given information. This principle indicates that the total mechanical energy of the system is the same after the object is launched as it is before the launch. The resulting equation will provide us with the value of k/m that we need in order to determine the angular frequency from k ω =. m SOLUTION The conservation of mechanical energy states that the final total mechanical energy E f is equal to the initial total mechanical energy E 0. The expression for the total mechanical energy for a spring/mass system is given by Equation 0.4, so that we have mv f + Iω f + mgh f + kx f = mv + Iω + mgh kx 0 E f Since the object does not rotate, the angular speeds ω f and ω 0 are zero. Since the object is initially at rest, the initial translational speed v 0 is also zero. Moreover, the motion takes place horizontally, so that the final height h f is the same as the initial height h 0. Lastly, the spring is unstrained after the launch, so that x f is zero. Thus, the above expression can be simplified as follows: k vf mv = kx or = f 0 m x Substituting this result into Equation 0. shows that 0 E 0 k v v.50 m/s ω = = = = = m x x m f f rad/s 38. REASONING As the sphere oscillates vertically, it is subject to conservative forces only: gravity and the elastic spring force. Therefore, its total mechanical energy E = mv + Iω + mgh+ ky (Equation 0.4) is conserved, and we use the variable y to denote the vertical stretching of the spring relative to its unstrained length. We will apply the energy conservation principle to find the spring constant k.

29 504 SIMPLE HARMONIC MOTION AND ELASTICITY SOLUTION In terms of the final and initial total mechanical energies E f and E 0, the conservation principle gives us the following starting point: mv + f Iω + mgh f f + ky f = mv + 0 Iω + mgh ky 0 E f E 0 () We note that the sphere does not rotate, so the angular speeds ω 0 and ω f are zero. Equation () then becomes mvf mghf ky f mv 0 mgh0 ky 0 Solving Equation () for the spring constant k, we obtain ( f 0) ( 0 f ) ( 0 f ) + + = + + () ( 0 f ) + ( 0 f ) mv v mgh h k y y = m v v + mg h h or k = yf y0 As the spring stretches from y 0 = 0. m to y f = 0.3 m, the sphere moves downward by y f y 0 = 0. m, so the difference between the sphere s initial and final heights is h0 hf = 0. m, which is positive since h 0 is greater than h f. Therefore, from Equation (3), the spring constant k is ( ) ( ) ( ) + ( )( ) 0.60 kg 5.70 m/s 4.80 m/s 0.60 kg 9.80 m/s (0. m) k = = 80 N/m ( 0.3 m) ( 0. m) (3) 39. REASONING AND SOLUTION a. Now look at conservation of energy before and after the split Before mv = ka Solving for the amplitude A gives max A= v max After m m v = v max = ka Solving for the amplitude A' gives Therefore, we find that A = v max m k m k A m A' = = = m

30 Chapter 0 Problems 505 Similarly, for the frequency, we can show that f ' = f = (3.00 Hz) = 4.4 Hz b. If the block splits at one of the extreme positions, the amplitude of the SHM would not change, so it would remain as m The frequency would be f ' = f = (3.00 Hz) = 4.4 Hz 40. REASONING The amount by which the spring stretches due to the weight of the.-kg object can be calculated using Equation 0. (with the variable x replaced by y), where the force F y Applied is equal to the weight of the object. The position of the object when the spring is stretched is the equilibrium position for the vertical harmonic motion of the objectspring system. SOLUTION a. Solving Equation 0. for y with F y Applied equal to the weight of the object gives Applied y mg F (. kg)(9.80 m/s ) y = = = = m k k 0 N/m b. The object is then pulled down another 0.0 m and released from rest ( v 0 = 0m/s). At this point the spring is stretched by an amount of m m = 0.9 m. We will let this point be the zero reference level ( h = 0 m) for the gravitational potential energy. The kinetic energy, the gravitational potential energy, and the elastic potential energy at the point of release are: KE (0 m/s) = mv = m = 0 J gravity 0 PE = mgh = mg(0 m) = 0 J elastic ky 0 PE = = (0 N/m)(0.9 m) = 5.0 J The initial total mechanical energy E 0 is the sum of these three energies, so E 0 = 5.0 J. When the object has risen a distance of h = 0.0 m above the release point, the spring is stretched by an amount 0.9 m 0.0 m = m. Since the total mechanical energy is conserved, its value at this point is still 5.0 J. Thus, E = KE + PE + PE gravity elastic

31 506 SIMPLE HARMONIC MOTION AND ELASTICITY E = mv + mgh+ ky v 5.0 J = (. kg) + (. kg)(9.80 m/s )(0.0 m) + (0 N/m)(0.090 m) Solving for v yields v =. m/s. 4. SSM REASONING Using the principle of conservation of mechanical energy, the initial elastic potential energy stored in the elastic bands must be equal to the sum of the kinetic energy and the gravitational potential energy of the performer at the point of ejection: k x = mv 0 + mgh where v 0 is the speed of the performer at the point of ejection and, from the figure at the right, h = x sin θ. Thus, k x = mv 0 + mgxsinθ () From the horizontal motion of the performer θ x h where v 0x = v 0 cosθ () v 0x = s x t (3) and s x = 6.8 m. Combining equations () and (3) gives Equation () becomes: v 0 = s x t(cos θ ) k x = m s x t cos θ + mgxsinθ This expression can be solved for k, the spring constant of the firing mechanism. SOLUTION Solving for k yields: s x k = m xt cosθ + mg(sinθ) x

32 Chapter 0 Problems m k = (70.0 kg) (3.00 m)(.4 s)(cos 40.0 ) ( )( )( ) 70.0 kg 9.80 m/s sin =.37 0 N/m 3.00 m 4. REASONING AND SOLUTION The bullet (mass m) moves with speed v, strikes the block (mass M) in an inelastic collision and the two move together with a final speed V. We first need to employ the conservation of linear momentum to the collision to obtain an expression for the final speed: mv mv = ( m + M ) V or V = m + M The block/bullet system now compresses the spring by an amount x. During the compression the total mechanical energy is conserved so that ( m + M) V = kx Substituting the expression for V into this equation, we obtain mv ( m + M) = kx m + M Solving this expression for v gives ( + ) ( 845 N/m)( 0.00 m) (.5 kg) m ( kg) kx m M v = = = 9 m/s 43. REASONING As the ball swings down, it reaches it greatest speed at the lowest point in the motion. One complete cycle of the pendulum has four parts: the downward motion in which the ball attains its greatest speed at the lowest point, the subsequent upward motion in which the ball slows down and then momentarily comes to rest. The ball then retraces its motion, finally ending up where it originally began. The time it takes to reach the lowest point is one-quarter of the period of the pendulum, or t = (/4)T. The period is related to the angular frequency ω of the pendulum by Equation 0.4, T = π/ω. Thus, the time for the ball to reach its lowest point is π t = T = 4 4 ω

33 508 SIMPLE HARMONIC MOTION AND ELASTICITY The angular frequency ω of the pendulum depends on its length L and the acceleration g due to gravity through the relation ω = g/ L (Equation 0.6). Thus, the time is π π π t = = = 4 ω 4 g L L g SOLUTION After the ball is released, the time that has elapsed before it attains its greatest speed is π L π 0.65 m t = = = 0.40 s g 9.80 m/s 44. REASONING The magnitude g of the acceleration due to gravity is related to the length L and frequency f of the simple pendulum by π f = g L (Equation 0.6). Squaring both sides of Equation 0.6 and solving for g, we obtain g 4 π f = or g = 4π f L () L The frequency f is the number of complete vibrations made per second. In measuring the frequency of the simple pendulum, the astronauts recorded N complete vibrations occurring over a total elapsed time t. The pendulum s oscillation frequency is, then, given by SOLUTION Substituting Equation () into Equation (), we obtain N f = () t ( ) ( ) ( 80 s) N 4π N L 4π 00. m L g = 4π = = = 6.0 m/s t t 45. REASONING a. The angular frequency ω of a simple pendulum can be found directly from Equation 0.6 as ω = g/ L, where g is the magnitude of the acceleration due to gravity and L is the length of the pendulum. b. The total mechanical energy of the pendulum as it swings back and forth is the gravitational potential energy it has just before it is released, since the pendulum is released from rest and has no initial kinetic energy. The reason is that friction is being neglected, and the tension in the cable is always perpendicular to the motion of the bob, so the tension does no work. Thus, the work done by nonconservative forces, such as friction and tension, is

34 Chapter 0 Problems 509 zero. This means that the total mechanical energy is conserved (see Equation 6.9b) and is the same at all points along the motion, including the initial point where the bob is released. c. To find the speed of the bob as it passes through the lowest point of the swing, we will use the conservation of energy, which relates the total mechanical energy at the lowest point to that at the highest point. SOLUTION a. The angular frequency of the pendulum is ω = g 9.80 m/s 3.5 rad/s L = 0.79 m = (0.6) b. At the moment the pendulum is released, the only type of energy it has is its gravitational potential energy. Thus, its potential energy PE is equal to its initial total mechanical energy E 0, so PE = E 0. According to Equation 6.5, the potential energy of the pendulum is PE = mgh, where m is the mass of the bob and h is its height above its equilibrium position (i.e., its position when the pendulum hangs straight down). The drawing shows that this height is related to the length L of the pendulum by h= L( cos8.50 ). Thus, the total mechanical energy of the pendulum is E = mgh= mgl( cos8.50 ) 0 ( )( )( )( ) = 0.4 kg 9.80 m/s 0.79 m cos8.50 =.0 0 J 8.50 L h c. As the bob passes through the lowest point of the swing, it has only kinetic energy, so its total mechanical energy is E = mv. Since the total mechanical energy is conserved ( E E ) = we have that mv = E f 0, Solving for the final speed gives f f f 0 ( ) E0.0 0 J vf = = = 0.4 m/s m 0.4 kg 46. REASONING AND SOLUTION Applying Equation 0.6 and recalling that frequency and period are related by f = /T, π π f = = T g L

35 50 SIMPLE HARMONIC MOTION AND ELASTICITY where L is the length of the pendulum. Thus, Solving for L gives T = π ( ) L g T 9. s L = g = 9.80 m/s = m π π 47. REASONING According to Equation 0.5, the angular frequency ω of a physical pendulum is ω = mgl / I and depends on the ratio of the mass m to the moment of inertia I. Since the moment of inertia is directly proportional to the mass (see Equation 9.6), the mass algebraically cancels. Thus, the angular frequency is independent of the mass of the physical pendulum. According to Equation 0.4, the period is T = π /ω. Since the angular frequency ω is independent of the mass, so is the period. These two expressions will allow us to determine the periods of the wood and metal pendulums. SOLUTION a. The period T of a pendulum is given by Equation 0.4 as T = π/ω, where ω is its angular frequency. The angular frequency of a physical pendulum is given by Equation 0.5 as ω = mgl / I, where m is its mass, L is the distance from the pivot to the center of mass, and I is the moment of inertia about the pivot. Combining these two relations yields π π T = = = π ω mgl I I mgl The moment of inertia of a meter stick (a thin rod) that is oscillating about an axis at one end is given in Table 9. as I = ml, where L 0 is the length of the stick. Since the meter 3 0 stick is uniform, the distance L from one end to its center of mass is L = L 0. Therefore, the period of oscillation of the wood pendulum is T ml 3 0 L0 I = π = π = π mgl mg ( L0 ) 3g ( ) = π.00 m =.64 s m /s ( ) b. The period is the same for the metal pendulum, since the mass has been eliminated algebraically in the expression for T.

36 Chapter 0 Problems REASONING For small-angle displacements, the frequency of simple harmonic motion for a physical pendulum is determined by π f = mgl/ I, where L is the distance between the axis of rotation and the center of gravity of the rigid body of moment of inertia I. Since the frequency f and the period T are related by f =/T, the period of pendulum A is given by T A = π I mgl Since the pendulum is made from a thin, rigid, uniform rod, its moment of inertia is given by I = (/3)md, where d is the length of the rod. Since the rod is uniform, its center of gravity lies at its geometric center, and L = d / Therefore, the period of pendulum A is given by T A = π d 3g For the simple pendulum we have T B = π d g SOLUTION The ratio of the periods is, therefore, T A = π d /(3g) = T B π d / g 3 = 0.86 π 49. REASONING The relation between the period T and angular frequency ω is T = ω mgl (Equation 0.6). The angular frequency of a physical pendulum is given by ω = I (Equation 0.5), where m is the mass of the pendulum, g is the acceleration due to gravity, L is the distance between the axis of rotation at the pivot point and the center of gravity of the rod, and I is the moment of inertia of the rod. According to Table 9., the moment of inertia of a thin uniform rod of length D is I = md. Combining these three equations 3 algebraically will give us an expression for the period that we seek. However, the length D of the rod is not given. Instead, the period of the simple pendulum is given. We will be able to use this information to eliminate the need for the missing length data. SOLUTION Substituting Equation 0.5 for ω into Equation 0.6, shows that the period of the physical pendulum is

37 5 SIMPLE HARMONIC MOTION AND ELASTICITY π π I T = = = π ω mgl mgl I Now we can use the expression I = md for the moment of inertia of the rod. In addition, 3 we recognize that the center of gravity of the uniform rod lies at the center of the rod, so that L= D. With these two substitutions the expression for the period becomes I 3 md D T = π = π = π () mgl mg D 3g ( ) At this point, we must deal with the unknown length D of the rod. To this end, we note that the period of the simple pendulum is given by Equations 0.6 and 0.6 as ω Simple π g D = = or T = π Simple T D g Simple Solving this expression for D/g and substituting the result into Equation () gives T = T = Simple ( 0.66 s) = 0.54 s REASONING As discussed in Section 0.4, a restoring torque τ is the cause of the backand-forth oscillatory motion. This restoring torque is given by τ k θ. This expression is analogous to Equation 0. for the Hooke s law restoring force. We will derive this expression in the present context, so that we can identify the constant k, which is analogous to the spring constant k in Equation 0.. Then, we will proceed by analogy with the simple harmonic motion that occurs for an object of mass m on a spring, in which case the angular frequency is given by ω = k / m (Equation 0.). SOLUTION As the object oscillates back and forth at the bottom of the bowl, it is subject to a restoring torque, which arises because of the object s weight. According to Equation 9., the magnitude of the torque τ is the magnitude mg of the weight times the lever arm l, so that τ = mgl. The minus sign is included because the torque is a restoring torque; it acts to reduce the angle θ (see the drawing that accompanies the problem statement). The lever arm in the text drawing is l = Rsinθ. The restoring torque on the object, then, is given by τ = mgrsinθ. When the angle θ is small enough so that the object oscillates in simple harmonic motion, we can replace sinθ with the angle θ itself, provided that the angle θ is expressed in radians. Thus, we have for small angles that τ mgr θ k

38 Chapter 0 Problems 53 The quantity mgr has a constant value k and is analogous to the spring constant k. As usual in rotational motion, the moment of inertia I appears in place of the mass m in equations. Thus, by analogy with ω = k / m (Equation 0.), we find that k mgr ω = = () I I The moment of inertia of the small object (assuming that all of its mass is located at the same distance R from the axis of rotation) is I = mr (Equation 9.6). Substituting this expression for I into Equation () gives ω = mgr = mgr or ω = g I mr R 5. SSM REASONING When the tow truck pulls the car out of the ditch, the cable stretches and a tension exists in it. This tension is the force that acts on the car. The amount ΔL that the cable stretches depends on the tension F, the length L 0 and cross-sectional area A of the cable, as well as Young s modulus Y for steel. All of these quantities are given in the statement of the problem, except for Young s modulus, which can be found by consulting Table 0.. ΔL SOLUTION Solving Equation 0.7, F = Y A L, for the change in length, we have 0 ( 890 N)( 9. m) FL0 4 Δ L = = = 5. 0 m AY π ( m) (.0 0 N /m ) 5. REASONING The stress in either cable is the ratio F/A of the magnitude F of the stretching force to the cross-sectional area A of the cable. The cables have circular cross-sections, so the area of each cable is given by A= πr. Therefore, the stress in either cable is given by F Stress = () π r We will use Equation () to solve for the stretching force F acting on the second cable. SOLUTION Setting the stresses in the cables equal via Equation (), and solving for the stretching force in the second cable gives

39 54 SIMPLE HARMONIC MOTION AND ELASTICITY F = F π r π r Stress in cable Stress in cable or F = F r r ( ) ( m) = 570 N ( = 70 N ) m 53. SSM REASONING AND SOLUTION According to Equation 0.0, it follows that ΔV P B ( ) V m Δ = =.6 0 N/m =.6 0 N/m.0 0 m (0.0) where we have expressed the volume V 0 of the cube at the ocean s surface as ( ) V 0 =.0 0 m =.0 0 m. Since the pressure increases by N/m per meter of depth, the depth is N/m = 60 m 4 N/m.0 0 m 54. REASONING The amount ΔL by which the bone changes length when a compression FL0 force or a tension force acts on it is specified by Δ L = (Equation 0.7), where F YA denotes the magnitude of either type of force, L 0 is the initial length of the bone, Y is the appropriate Young s modulus, and A is the cross-sectional area of the bone. The values of Young s modulus are given in Table 0. (Y Compression = N/m and Y Tension = N/m ). The values for F, L 0, and A are not given, but it is important to recognize that these variables have the same values for both types of forces. We will apply Equation 0.7 twice, once for the compression force and once for the tension force. Since F, L 0, and A have the same values in both of the resulting equations, we will be able to eliminate them algebraically and determine the amount ΔL Tension by which the bone stretches. SOLUTION Applying Equation 0.7 for both types of forces gives FL FL Δ L = and Δ L = Tension Y A Y A 0 0 Compression Tension Compression Dividing the left-hand equation by the right-hand equation and eliminating the common variables algebraically shows that

40 Chapter 0 Problems 55 Solving for ΔL Tension, we find that FL Δ L Y A Tension Δ L FL Compression 0 Y A Tension = = 0 Compression Y Compression Y 9 Compression N/m 5 5 Tension Compression 0 (.7 0 L L m ).6 0 Δ = Δ = = m Y.6 0 N/m Tension Y Tension 55. REASONING The change ΔL in the length of the rope is given by L FL ( YA) Δ = (see Equation 0.7), where F is the magnitude of the stretching force, L 0 is the unstretched length of the rope, A is its cross-sectional area, and Y is Young s modulus for nylon (see Table 0.). All the variables except F are known. According to Newton s third law, the action-reaction law, the force exerted on the rope by the skier is equal in magnitude to the force exerted on the skier by the rope. The force exerted on the skier by the rope can be obtained from Newton s second law, since the mass and acceleration of the skier are known. SOLUTION The change ΔL in the length of the rope is FL Δ L = 0 (0.7) YA Two horizontal forces act on the skier: () the towing force (magnitude = F) and () the resistive force (magnitude = f ) due to the water. The net force acting on the skier has a magnitude of F f. According to Newton s second law (see Section 4.3), this net force is equal to the skier s mass m times the magnitude a of her acceleration, or F f = ma Solving this equation for F and substituting the result into Equation 0.7, we find that ( + ) 30 N ( 59 kg)( 0.85 m/s ) + ( m) YA ( 9 )( 5 ) FL0 f ma L0 Δ L = = = =.9 0 m YA N/m.0 0 m We have taken the value of Y = N/m for nylon from Table REASONING When the object is placed on top of the pipe, the pipe is compressed due to the weight of the object. The amount ΔL of compression depends on the weight F, the length L 0 and cross-sectional area A of the pipe, as well as Young s modulus Y for steel. All 0 /

41 56 SIMPLE HARMONIC MOTION AND ELASTICITY of these quantities, except for the weight, are given in the statement of the problem. Young s modulus can be found by consulting Table 0.. SOLUTION Using Equation 0.7, we find that the weight F of the object is Δ L ( ) 7 m ( ) 4 F = Y A=.0 0 N /m π 65 0 m = 4. 0 N L0 3.6 m Note that in this calculation we have used the fact that the circular cross section of the pipe is a circle of area A= πr. 57. REASONING The shear stress is equal to the magnitude of the shearing force exerted on the bar divided by the cross sectional area of the bar. The vertical deflection ΔY of the right end of the bar is given by Equation 0.8 [ F = S(ΔY /L 0 )A]. SOLUTION a. The stress is F A = mg A = (60 kg)(9.80 m/s ) m = N/m b. Taking the value for the shear modulus S of steel from Table 0., we find that the vertical deflection ΔY of the right end of the bar is ΔY = F L 0 A S = ( N/m 0.0 m ) N/m = m 58. REASONING AND SOLUTION F = S(ΔX/L 0 )A for the shearing force. The shear modulus S for copper is given in Table 0.. From the figure we also see that tan θ = (ΔX/L 0 ) so that 6 F N θ = tan tan = = 0.09 SA 0 ( 4. 0 N/m )( m ) 59. SSM REASONING AND SOLUTION The shearing stress is equal to the force per unit area applied to the rivet. Thus, when a shearing stress of Pa is applied to each rivet, the force experienced by each rivet is 8 F = (Stress) A = (Stress)( πr ) = (5.0 0 Pa) π œ3 4 [ (5.0 0 m) ] = N

42 Chapter 0 Problems 57 Therefore, the maximum tension T that can be applied to each beam, assuming that each rivet carries one-fourth of the total load, is 5 4F =.6 0 N. 60. REASONING Both cylinders experience the same force F. The magnitude of this force is related to the change in length of each cylinder according to Equation 0.7: F = Y(ΔL / L 0 )A. See Table 0. for values of Young s modulus Y. Each cylinder decreases in length; the total decrease being the sum of the decreases for each cylinder. SOLUTION The length of the copper cylinder decreases by ΔL copper = FL 0 YA = FL 0 Y(πr ) = (6500 N)(3.0 0 m) (. 0 N/m )π(0.5 0 m) = m Similarly, the length of the brass decreases by ΔL brass = (6500 N)(5.0 0 m) ( N/m )π (0.5 0 m) = m Therefore, the amount by which the length of the stack decreases is m. 6. REASONING AND SOLUTION Equation 0.0 gives the desired result. Solving for ΔV / V 0 and taking the value for the bulk modulus B of aluminum from Table 0.3, we obtain ΔV = ΔP V 0 B =.0 05 Pa = N/m 6. REASONING It takes force to stretch the wire. This force arises because the tuning peg at one end of the wire pulls on the fixed support at the other end. In accord with Newton s action-reaction law, the fixed support pulls back. As a result of the oppositely-directed pulling forces at either end of the wire, the wire experiences an increased tension. For each turn, the change in length of the wire is equal to the circumference of the tuning peg, ΔL = π r p, where r p is the radius of the tuning peg. This change in length is related to the tension by virtue of Young s modulus Y for steel, which has a value of Y =.0 0 N/m. SOLUTION The tension is the force F that is required to stretch the wire (unstrained length = L 0, cross-sectional area = A) by an amount ΔL and is determined by Young s modulus according to Equation 0.7:

43 58 SIMPLE HARMONIC MOTION AND ELASTICITY ΔL F = Y A L 0 Assuming that the wire has a circular cross-section, A= π r w, where r w is the radius of the wire. When the tuning peg is turned through two revolutions, the length of the wire will increase by an amount equal to twice the circumference of the peg. Thus, Δ L = ( π r ), where r p is the radius of the tuning peg. With these substitutions, Equation 0.7 becomes: p 4π r p 4Yr F = Y πr = r L 0 L0 ( π ) p w w (.0 0 N/m )(.8 0 m) F = π ( m) =. 0 N 0.76 m 63. REASONING AND SOLUTION From the drawing we have Δx = m and We now have Stress = F/A. Therefore, A = πrδx = π(.00 0 m)( m) F = (Stress)A = ( Pa)[π(.00 0 m)( m)] = N 64. REASONING The person is to be suspended by N pieces of mohair, which collectively support the person against the downward force W = mg of gravity, where m = 75 kg is the mass of the person and g is the magnitude of the acceleration due to gravity. Therefore, the tension T in each piece must be T = mg N, and the total number of pieces is The tension T in one piece of mohair is given by mg N = () T ΔL ΔL T = Y A= Y r L L π 0 0 (0.7) where Y is Young s modulus, ΔL/L 0 is the strain, and A = π r is the cross-sectional area of a single piece of mohair with a radius r.

44 Chapter 0 Problems 59 SOLUTION We take the value of Young s modulus 9 Y =.9 0 N/ m for mohair from Table 0. in the text. Substituting Equation 0.7 into Equation (), we obtain ( 75 kg)( 9.80 m/s ) 9 6 (.9 0 N/m )( 0.00) π ( 3 0 m) mg N = = = Δ L Y πr L SSM REASONING Our approach is straightforward. We will begin by writing Δ L Equation 0.7 F = Y A as it applies to the composite rod. In so doing, we will L 0 use subscripts for only those variables that have different values for the composite rod and the aluminum and tungsten sections. Thus, we note that the force applied to the end of the composite rod (see Figure 0.7) is also applied to each section of the rod, with the result that the magnitude F of the force has no subscript. Similarly, the cross-sectional area A is the same for the composite rod and for the aluminum and tungsten sections. Next, we will express the change ΔL Composite in the length of the composite rod as the sum of the changes in lengths of the aluminum and tungsten sections. Lastly, we will take into account that the initial length of the composite rod is twice the initial length of either of its two sections and thereby simply our equation algebraically to the point that we can solve it for the effective value of Young s modulus that applies to the composite rod. SOLUTION Applying Equation 0.7 to the composite rod, we obtain Δ L Composite F = Y A Composite L 0, Composite () Since the change ΔL Composite in the length of the composite rod is the sum of the changes in lengths of the aluminum and tungsten sections, we have ΔL Composite = ΔL Aluminum + ΔL Tungsten. Furthermore, the changes in length of each section can be expressed using Equation 0.7 FL L 0 Δ =, so that YA FL FL 0, Aluminum 0, Tungsten Δ L =Δ L +Δ L = + Composite Aluminum Tungsten Y A Y A Substituting this result into Equation () gives Aluminum Tungsten

45 50 SIMPLE HARMONIC MOTION AND ELASTICITY Y A Y A FL FL Composite Composite 0, Aluminum 0, Tungsten F = Δ L = + Composite L L Y A Y A 0, Composite 0, Composite Aluminum Tungsten L 0, Aluminum 0, Tungsten = Y + Composite L Y L Y 0, Composite Aluminum 0, Composite Tungsten In this result we now use the fact that L 0, Aluminum /L 0, Composite = L 0, Tungsten /L 0, Composite = / and obtain = Y + Composite Y Y Aluminum Tungsten Solving for Y Composite shows that Y L 0 ( )( ) ( N/m ) ( N/m ) Y Y N/m N/m Tungsten Aluminum = = =. 0 N/m Composite 0 Y + Y Tungsten Aluminum + The values for Y Tungsten and Y Aluminum have been taken from Table REASONING Table 0.4 indicates that the shear stress is given by F/A, and the shear strain is given by ΔX/L 0, where A is the area of the plate s square face and L 0 is the plate s thickness. The maximum amount of force F that can be applied as in the referenced text figure is determined by the static frictional force between the plate and the surface on which MAX it rests. The maximum static frictional force is given by fs = µ sfn (Equation 4.7), where µ s is the coefficient of static friction and F N is the normal force that presses together the plate and the surface on which it rests. Since the plate does not accelerate in the direction perpendicular to the surface, the normal force and the plate s weight balance, so that FN = mg. Thus, we know that MAX s µ s N µ s F = f = F = mg SOLUTION a. Referring to Table 0.4 for the definition of shear stress, we find that ( )( ) ( m) Maximum possible F µ kg 9.80 m/s smg = = = = shear stress A A 70 N/m b. Referring to Table 0.4 for the definition of shear strain and recognizing that the shear modulus S is given, we find that

46 Chapter 0 Problems 5 0 Maximum possible shear stress 70 N/m Maximum possible ΔX = = = = shear strain L S.0 0 N/m c. The maximum possible amount of shear deformation ΔX can now be determined from the result in part b as follows: ( )( ) Maximum possible 8 0 shear strain L m m Δ X = = = 67. REASONING AND SOLUTION a. Strain = ΔL/L 0 = F/(YA). In this case the area subjected to the compression is given by ( ) ( ) ( ) A= π rout rin = π =.64 0 m 3 4 and the force is F = mg. Taking the value for Young s modulus Y for bone compression from Table 0., we find that ( 63 kg)( 9.80 m/s ) 9 4 ( Pa)(.64 0 m ) Strain = =.5 0 b. ΔL = Strain L 0 = ( )(0.30 m) = m 68. REASONING We will use energy conservation. The person falls from rest and does not rotate, so initially he has only gravitational potential energy. Ignoring air resistance and friction, we may apply the conservation of mechanical energy. Since the person strikes the ground stiff-legged and comes to a halt without rotating, all of the energy he had to begin with must be absorbed by his leg as elastic potential energy. The height through which he falls determines the amount of his gravitational potential energy and, hence, the amount of energy his leg must absorb. According to Equation 0.3, the elastic potential energy of an ideal spring is PE k x =, so we will need k Effective for a thigh bone. To find it, we consider Elastic Effective Equation 0. for the applied force needed to change the length of an ideal spring: F x Applied = k effective x, where k effective is the spring constant and x is the displacement. To change the length of a bone, in comparison, the necessary applied force is given by Equation 0.7 as follows: 4

47 5 SIMPLE HARMONIC MOTION AND ELASTICITY F Applied x = Y Δ L L 0 A = ( YA / L 0 ) The effective spring constant k Effective Δ L The change in length or the displacement x In this expression Y is Young s modulus, A is the effective cross-sectional area of the bone, L 0 is the initial length of the bone, and ΔL is the change in length. Associating ΔL with x, we see that the effective spring constant of the bone is given by k Effective YA = () L Equation 0.3, which specifies the elastic potential energy of an ideal spring as PE k x =, contains the displacement x of the spring. To eliminate it, we turn to Elastic Effective Equation 0., which gives the applied force as F x Applied = k Effective x. Solving Equation 0. for x and substituting the result into Equation 0.3 gives PE Applied F x = k x = k = k Effective Elastic Effective Effective 0 ( F ) Applied x k Effective SOLUTION The conservation of mechanical energy states that the final total mechanical energy E f is equal to the initial total mechanical energy E 0. The expression for the total mechanical energy for a spring/object system is given by Equation 0.4, so we have () mv f + Iω f + mgh f + kx f = mv + Iω + mgh kx 0 E f Since the person does not rotate, the angular speeds ω f and ω 0 are zero. The person is at rest both initially and finally, so the initial and final translational speeds v 0 and v f are also zero. Moreover, the thighbone is initially unstrained, with the result that x 0 is zero. Thus, the above expression can be simplified to give mgh + kx = mgh f f 0 Using Equation () to express the final elastic potential energy of the thighbone, we can write the simplified energy-conservation equation as follows: Applied Applied ( F ) ( F ) E 0 x or f 0 0 f keffective mgh + = mgh h h = mgk x Effective As the man falls, his center of gravity moves from its initial height of h 0 to its final height of h f, which is a distance of h 0 h f. Using Equation () for the effective spring constant of the bone, we find

48 Chapter 0 Problems 53 Applied Applied ( ) ( ) Effective x x 0 h h = F = F L 0 f mgk mgya 4 ( N) ( 0.55 m) 9 4 ( )( )( )( ) = = 65 kg 9.80 m/s N/m m 0.56 m 69. SSM REASONING AND SOLUTION Strain = ΔL/L 0 = F/(YA) where F = mg and A = π r. Setting the strain for the spider thread equal to the strain for the wire Thus, F = Y A Spider thread F Y A Aluminum wire so that FYr r = FY F Y r = F Y Taking the value for Young s modulus Y for aluminum from Table 0., we find that 9 6 ( 95 kg)( 9.80 m/s )( Pa)( 3 0 m) 3 0 (.0 0 kg)( 9.80 m/s )( Pa) r = =.0 0 m 70. REASONING We will find displacement ΔX of the upper surface relative to the lower ΔX surface from F = S A (Equation 0.8), where F is the applied horizontal force (see L 0 Figure 0.30), S is the shear modulus of brass, L 0 is the distance between the upper and lower surfaces, and A is the cross-sectional area of either surface. Solving Equation 0.8 for ΔX yields ΔX = FL 0 SA = F S Same for all orientations L 0 A r Depends upon orientation 3 () Note that the term L 0 /A in Equation () depends on the orientation of the block. The block can rest on any one F L 0 = m 0.00 m 0.00 m

49 54 SIMPLE HARMONIC MOTION AND ELASTICITY of its three unique surfaces. To make the displacement ΔX of the top surface as large as possible, Equation () shows that we must orient the block so as to make the ratio L 0 /A as large as possible. The largest possible value for the numerator of this ratio is the longest side of the block, so we choose L 0 = m. Gluing the block to the table so that the four longest edges are vertical means that the upper surface of the block is the surface with the smallest area: A = 0.00 m 0.00 m (see the drawing). This combination of the largest possible value of L 0 and the smallest possible value of A makes the ratio L 0 /A as large as possible. Therefore, this is the orientation with the greatest possible displacement ΔX of the upper surface relative to the lower surface when the horizontal force is applied to the upper surface. 0 SOLUTION We obtain the shear modulus S = N/m of brass from Table 0. in the text. From Equation (), the maximum possible displacement of the upper surface relative to the lower surface is ( 770 N)( m) 0 ( N/m )( 0.00 m)( 0.00 m) FL0 6 Δ X = = = m SA 7. REASONING The unstretched length L 0 of the cable can be found from the relation L = YA( Δ L) F (Equation 0.7), where Y is Young s modulus for steel (see Table 0.), 0 / A is the cross-sectional area of the cable, ΔL is the amount by which it stretches, and F is the magnitude of the stretching force. All the variables except F are known. According to Newton s third law, the action-reaction law, the force exerted on the cable by the skier has the same magnitude as the force exerted on the skier by the cable. The force exerted on the skier by the cable can be obtained from Newton s second law, since the mass and acceleration of the skier are known. SOLUTION The unstrained length of the cable is L 0 YA( ΔL) = (0.7) F To determine F, we examine the following free-body diagram of the skier. For convenience, the +x direction is taken to be parallel to the slope and to point upward (see the drawing). F +x mg sin f W = mg Free-body diagram for the skier W = mg

50 Chapter 0 Problems 55 Three forces act on the skier in the x direction: () the towing force (magnitude = F), () the frictional force (magnitude = f ) exerted on the skis by the snow, and (3) the component of the skier s weight that is parallel to the x axis (magnitude = W sin = mg sin ). This component is shown to the right of the free-body diagram. The net force acting on the skier has a magnitude of F f mgsin. According to Newton s second law (see Section 4.3), this net force is equal to the skier s mass times the magnitude of her acceleration, or F f mgsin = ma Solving this equation for F and substituting the result into Equation 0.7, we find that L ( ) ( )( 5 )( 4 Δ L.0 0 N/m m.0 0 m) ( )( ) ( ) ( )( ) YA = = = m ma+ f + mgsin 6 kg. m/s + 68 N + 6 kg 9.80 m/s sin 0 We have taken the value of Y =.0 0 N/m for steel from Table REASONING The unstrained length L 0 of the cord is related to the stretching force (tension) F on the cord, the amount ΔL by which the cord is stretched, the cord s cross-sectional area A, and Young s modulus Y = N/m (see Table 0. in the text) by ΔL F = Y A (0.7) L 0 While swinging at a speed v on the end of the cord, the bowling ball (mass = m) follows a circular path, and the tension F in the cord must be sufficient to supply the necessary mv centripetal force Fc = (Equation 5.3). Ignoring the stretch in the cord, the radius r of r the circle that the bowling ball traverses is equal to the length L 0 of the cord. Thus, Equation 5.3 becomes F c 0 mv = (5.3) L At the lowest point, the upward force F on the ball is opposed by the downward weight force mg, so the difference between these two forces equals the centripetal force F c : mv mv Fc = F mg or F mg L = = L + () 0 0 To find the speed of the bowling ball, we take advantage of energy conservation. As the ball swings downward from release to its lowest point, it is subject to only one nonconservative force: the tension in the cord. This force is perpendicular to the bowling ball s velocity at every instant, and, therefore, does no work. Consequently, the total mechanical energy

51 56 SIMPLE HARMONIC MOTION AND ELASTICITY ω E = mv + I + mgh+ kx (Equation 0.4) of the bowling ball is conserved. The bowling ball starts from rest and undergoes no rotation (ω 0 = ω f = 0 rad/s), so its initial translational kinetic energy is zero, as are its initial and final rotational kinetic energies. The cord is initially unstretched, and its final stretch x = ΔL is negligible, so the initial and final elastic potential energies are both zero. When we apply the energy conservation principle, then, we obtain mv f E f + mgh = mgh f 0 Solving Equation () for the square of the speed we find that m v f + mghf E 0 () v f of the bowling ball at its lowest point, = m gh or v = g ( h h ) (3) 0 f 0 f We will use Equations (0.7), (), and (3) to solve for the unstretched length L 0 of the cord. SOLUTION Because Equation () refers to the instant when the bowling ball reaches its lowest point, the speed v in Equation () is identical with the final speed v f in Equation (3): v = v f. Making this substitution in Equation (), and setting Equation (0.7) and Equation () equal, we obtain f ΔL mv Y A= + mg L 0 L0 Multiplying both sides of Equation (4) by L 0 and solving, we find (4) ΔL Y AL L 0 0 mvf = L0 L0 ( Δ ) Y L A mvf + mgl0 or L0 = (5) mg Substituting Equation (3) for the square of the final speed into Equation (5) yields the unstrained length of the cord: ( Δ ) ( ) ( Δ ) Y L A mg h h Y L A L h h mg mg ( ) 0 f 0 = = 0 f ( N/m )(.7 0 m)( m ) ( 6.8 kg)( 9.80 m/s ) ( ) =.4 m =.3 m 73. SSM REASONING Equation 0.0 can be used to find the fractional change in volume of the brass sphere when it is exposed to the Venusian atmosphere. Once the fractional change in volume is known, it can be used to calculate the fractional change in radius.

52 Chapter 0 Problems 57 SOLUTION According to Equation 0.0, the fractional change in volume is ΔV = ΔP V 0 B = Pa Pa = Here, we have used the fact that Δ P = Pa.0 0 Pa = Pa, and we have taken the value for the bulk modulus B of brass from Table 0.3. The initial volume of the sphere is 4 3 V = π r. If we assume that the change in the radius of the sphere is very small relative to the initial radius, we can think of the sphere's change in volume as the addition or subtraction of a spherical shell of volume ΔV, whose radius is r 0 and whose thickness is Δr. Then, the change in volume of the sphere is equal to the volume of the shell and is given by ( 4π ) 0 Δ r/ r, we have 0 Therefore, Δ V = r Δ r. Combining the expressions for V 0 and ΔV, and solving for Δr ΔV = r 3 V 0 0 ( ) Δ r = = r REASONING The number of times the diaphragm moves back and forth is the frequency f of the motion (in cycles/s or Hz) times the time interval t. The frequency f is related to the angular frequency ω (in rad/s) by Equation 0.6 (ω = πf ). SOLUTION Solving Equation 0.6 (ω = πf ) for the frequency f gives ω f = π The number of times the diaphragm moves back and forth in.5 s is 4 ω rad/s Number of times = ft = t = (.5 s) = π π REASONING The amplitude of simple harmonic motion is the distance from the equilibrium position to the point of maximum height. The angular frequency ω is related to the period T of the motion by Equation 0.6. The maximum speed attained by the person is the product of the amplitude and the angular speed (Equation 0.8). SOLUTION

53 58 SIMPLE HARMONIC MOTION AND ELASTICITY a. Since the distance from the equilibrium position to the point of maximum height is the amplitude A of the motion, we have that A = 45.0 cm = m. b. The angular frequency is inversely proportional to the period of the motion: π π ω = = = 3.3 rad /s (0.6) T.90 s c. The maximum speed v max attained by the person on the trampoline depends on the amplitude A and the angular frequency ω of the motion: ( )( ) vmax = Aω = m 3.3 rad /s =.49 m/s (0.8) 76. REASONING The length L of a simple pendulum is related to its frequency f via g π f = (Equation 0.6). In terms of its period T, the frequency of a simple pendulum L is f = (Equation 0.5), so we have π g =. Solving for the length L, we obtain T T L T L or T L or L T g π = g 4π = g = 4π () We will use Equation () to calculate the difference L L between the final and initial lengths of the pendulum. SOLUTION The period of the original pendulum is T =.5 s. When its length is increased from L to L, its period increases by 0.0 s to T =.5 s s =.45 s. From Equation (), the difference L L between the pendulum s final and initial lengths is T g T g g 9.80 m/s ( ) ( ) ( ) L L = = T T =.45 s.5 s = 0.3 m 4π 4π 4π 4π 77. SSM REASONING Each spring supports one-quarter of the total mass m total of the system (the empty car plus the four passengers), or 4 m total. The mass m one passenger of one passenger is equal to 4 m total minus one-quarter of the mass m empty car of the empty car: one passenger 4 total 4 empty car m = m m ()

54 Chapter 0 Problems 59 The mass of the empty car is known. Since the car and its passengers oscillate up and down in simple harmonic motion, the angular frequency ω of oscillation is related to the spring constant k and the mass m supported by each spring by: Solving this expression for Equation () gives 4 total k ω = (0.) m 4 total 4 m total ( m 4 total k/ ω ) m one passenger = and substituting the result into k = m 4 empty car () ω The angular frequency ω is inversely related to the period T of oscillation by ω = π /T (see Equation 0.4). Substituting this expression for ω into Equation () yields m one passenger k = π T 4 m empty car SOLUTION The mass of one of the passengers is m one passenger N/m m 4 empty car 4 k = = ( 560 kg) = 6 kg π π T s 78. REASONING a. According to the discussion in Section 0.8, the stress is the magnitude of the force per unit area required to cause an elastic deformation. We can determine the maximum stress that will fracture the femur by dividing the magnitude of the compressional force by the cross-sectional area of the femur. b. The strain is defined in Section 0.8 as the change in length of the femur divided by its original length. Equation 0.7 shows how the strain ΔL/L 0 is related to the stress F/A and Young s modulus Y (Y = N/m for bone compression, according to Table 0.). SOLUTION a. The maximum stress is equal to the maximum compressional force divided by the crosssectional area of the femur:

55 530 SIMPLE HARMONIC MOTION AND ELASTICITY F N Maximum stress = = =.7 0 N/m A m b. The strain ΔL/L can be found by rearranging Equation 0.7: 4 8 ( ) =.8 0 ΔL = F L 0 Y A = N/m.7 08 N/m Strain Stress 79. REASONING The applied force required to stretch the bow string is given by Equation Applied 0. as Fx = k x, where k is the spring constant and x is the displacement of the string. Solving for the displacement gives x= F Applied / k. SOLUTION The displacement of the bow string is Applied F x x + 40 N x = = = m k 480 N/m 80. REASONING The distance ΔX that the top surface of the disc moves relative to the bottom surface is given by Δ X = FL0 / ( SA) (see Equation 0.8), where F is the magnitude of the shearing force, L 0 is the thickness of the cartilage, A is the cross-sectional area, and S is the shear modulus. Since the cross-section is circular, A= πr, where r is the radius. SOLUTION The distance (shear deformation) ΔX is ( )( 3 ) FL0 N m 6 Δ X = = =.3 0 m (0.8) SA ( 7 ) (. 0 N/m π m) 8. REASONING Since the surface is frictionless, we can apply the principle of conservation of mechanical energy, which indicates that the total mechanical energy of the spring/mass system is the same at the x = 0 m v m m

56 Chapter 0 Problems 53 instant the block contacts the bottle (the final state of the system) and at the instant shown in the drawing (the initial state). Kinetic energy mv is one part of the total mechanical energy, and depends on the mass m and the speed v of the block. The dependence of the kinetic energy on speed is critical to our solution. In order for the block to knock over the bottle, it must at least reach the bottle. When launched with the minimum speed v 0 shown in the drawing, the block will reach the bottle with a final speed of v f = 0 m/s. We will obtain the desired initial speed v 0 by solving the energy-conservation equation for this variable. SOLUTION The conservation of mechanical energy states that the final total mechanical energy E f is equal to the initial total mechanical energy E 0. The expression for the total mechanical energy for a spring/mass system is given by Equation 0.4, so that we have mv f + Iω f + mgh f + kx f = mv + Iω + mgh kx 0 E f Since the block does not rotate, the angular speeds ω f and ω 0 are zero. Moreover, the block reaches the bottle with a final speed of v f = 0 m/s when the block is launched with the minimum initial speed v 0. In addition, the surface is horizontal, so that the final and initial heights, h f and h 0, are the same. Thus, the above expression can be simplified as follows: kx = kx + mv f 0 0 In this result, we are given no values for the spring constant k and the mass m. However, we are given a value for the angular frequency ω. This frequency is given by Equation 0. k ω =, which involves only the ratio k/m. Therefore, in solving the simplified energyconservation expression for the speed v 0 m, we will divide both sides by m, so that the ratio k/m can be expressed using Equation 0.. Substituting f 0 0 k = or v = x x 0 f 0 E 0 kx kx + mv m m m k ω = from Equation 0., we find m ( ) ( ) ( ) ( ) ( ) v = ω x x = 7.0 rad/s m m = 0.44 m/s 0 f 0 8. REASONING a. As the block rests stationary in its equilibrium position, it has no acceleration. According to Newton s second law, the net force acting on the block is, therefore, zero. This means that the downward-directed weight of the block must be balanced by an upward-directed force. This upward force is the restoring force of the spring and is produced because the spring is compressed. The compression must be enough for the spring to exert on the block

57 53 SIMPLE HARMONIC MOTION AND ELASTICITY a restoring force that has a magnitude equal to the block s weight. This balancing of forces will allow us to determine the magnitude of the spring s compression. b. As the block falls downward after being released, its speed is changing in the manner characteristic of simple harmonic motion. The block is not in equilibrium, and the forces acting on it do not balance to zero. Instead of thinking about forces, we may think about mechanical energy and its conservation. When the block is released from rest, the energy of the spring/block system is all in the form of gravitational potential energy. Being at rest, the block has no initial kinetic energy. It also has no initial elastic potential energy, since the spring is unstrained initially. When the block comes to a momentary halt at the lowest point in its fall, the energy is all in the form of elastic potential energy. Since the block is again at rest, it again has no kinetic energy. The spring has been compressed, and gravitational potential energy has been converted entirely into elastic potential energy. The amount by which the spring is compressed is determined by the amount of gravitational potential energy that must be converted into elastic potential energy. The amount must be enough that the elastic potential energy equals the gravitational potential energy. Thus, we will use energy conservation to determine the magnitude of the spring s compression. The compression of the spring is greater in the non-equilibrium case than in the equilibrium case. The reason is that in the non-equilibrium case, the block has been allowed to move, and its inertia carries it beyond its stationary equilibrium position on the spring. The compression of the spring must increase beyond that for the stationary equilibrium position in order to produce the force that is needed to decelerate the block to a momentary halt. SOLUTION a. As the block rests stationary on the spring, the downward-directed weight balances the upward-directed restoring force from the spring. The magnitude of the weight is mg, and the magnitude of the restoring force is given by Equation 0. without the minus sign as kx. Thus, we have mg = kx Magnitude of the weight Magnitude of the spring force or x = mg k ( )( 9.80 m/s ) 0.64 kg = 70 N/m = m b. The conservation of mechanical energy states that the final total mechanical energy E f is equal to the initial total mechanical energy E 0. The expression for the total mechanical energy for an object on a spring is given by Equation 0.4, so that we have mv f + Iω f + mgh f + kx f = mv + Iω + mgh kx 0 E f The block does not rotate, so the angular speeds ω f and ω 0 are zero. Since the block comes to a momentary halt on the spring and is released from rest, the translational speeds v f and v 0 are also zero. Because the spring is initially unstrained, the initial displacement x 0 of the spring is likewise zero. Thus, the above expression can be simplified as follows: E 0

58 Chapter 0 Problems 533 ( ) mgh + kx = mgh or kx = mg h h f f 0 f 0 f The term h 0 h f is the amount by which the spring has compressed, or h 0 h f = x f. Making this substitution into the simplified energy-conservation equation gives Solving for x f, we find ( ) kx = mg h h = mgx or kx = mg f 0 f f f ( )( ) mg 0.64 kg 9.80 m/s x = = = m k 70 N/m As expected, the spring compresses more in the non-equilibrium situation. 83. REASONING Since air resistance is being ignored and the bungee cord is assumed to be an ideal spring, the bungee jumper oscillates up and down in simple harmonic motion. The angular frequency ω of oscillation is related to the spring constant k and the mass m of the jumper by Equation 0.: ω = k/ m. Solving for the spring constant gives k = mω () The angular frequency ω is inversely proportional to the period T of oscillation according to ω = π /T (see Equation 0.4). Substituting this expression for ω into Equation () yields π k mω m T = = The mass of the jumper is known. The period of oscillation can be obtained from the fact that the jumper makes two complete oscillations in a time of 9.6 s. SOLUTION Since the bungee jumper moves through complete cycles in 9.6 s, the time to complete one cycle (the period of oscillation) is T = ( 9.6 s) = 4.8 s. The spring constant of the bungee cord is π π k = m = ( 8 kg) = 40 N/m T 4.8 s 84. REASONING The two blocks and the spring between them constitute the system in this problem. Since the surface is frictionless and the weights of the blocks are balanced by the normal forces from the surface, no net external force acts on the system. Thus, the system s total mechanical energy and total linear momentum are each conserved. The conservation of these two quantities will give us two equations containing the two unknown speeds with which the blocks move away. Using these equations, we will be able to obtain the speeds. SOLUTION The conservation of mechanical energy states that the final total mechanical energy E f is equal to the initial total mechanical energy E 0. Only the translational kinetic

59 534 SIMPLE HARMONIC MOTION AND ELASTICITY energies of the blocks and the elastic potential energy of the spring are of interest here. There is no rotational kinetic energy since there is no rotation. Gravitational potential energy plays no role, because the surface is horizontal and the vertical height does not change. Thus, the expression for the conservation of the total mechanical energy is m v + m v + kx f f f = m v + m v + kx E f E 0 Since the blocks are initially at rest, the initial translational speeds v 0 and v 0 are zero. In addition, the spring is neither compressed nor stretched after it is released, so that x f = 0 m. Thus, the above expression can be simplified as follows: mv + mv = kx () f f 0 Remembering that linear momentum is mass times velocity, we can express the conservation of the total linear momentum of the system as follows: m v f + m v = m v + m v f 0 0 Final total momentum Initial total momentum Since the blocks are initially at rest, v 0 and v 0 are zero, so that the expression for the conservation of linear momentum becomes mv mv v Substituting this result into Equation () gives mv f + = 0 or = () f f f m mv + mv m = kx f f 0 m Solving for v f, which we define to be the final speed of the.-kg block, shows that v f ( + ) (.7 kg)( 330 N/m)( 0.4 m) (. kg)(.7 kg +. kg) = mkx 0.5 m/s m m m = = Substituting this value into Equation () gives v mv (. kg)(.5 m/s) f = = = f m.7 kg m/s The speed of the.7-kg block is the magnitude of this result or m/s. 85. REASONING According to Equations 0.6 and 0., the frequency f at which an object k of mass m oscillates on a spring with a spring constant k is f = π m. We will apply this

60 Chapter 0 Problems 535 expression to the case when only the first object is attached to the spring and then again to the case when both objects are attached. SOLUTION For the two cases, we have f = k π m Only first object attached to spring and f = k π m + m Both objects attached to spring Dividing the expression for f by the expression for f gives k π m + + = = or = = + k f f m m f m m m f m m m π m + m Solving the result above on the right for the ratio m /m gives m m (.0 Hz) ( 4.00 Hz) f f = = = REASONING The strain in the wire is given by ΔL/ L 0. From Equation 0.7, the strain is therefore given by ΔL L 0 = F YA () where F must be equal to the magnitude of the centripetal force that keeps the stone moving in the circular path of radius R. Table 0. gives the value of Y for steel. SOLUTION Combining Equation () with Equation 5.3 for the magnitude of the centripetal force, we obtain ΔL F = L 0 Y(π r ) = (mv / R) Y (π r ) = (8.0 kg)( m/s) /(4.0 m) (.0 0 Pa) π( = m) 87. REASONING AND SOLUTION Use conservation of energy to find the speed of point A (take the pivot to have zero gravitational PE). E up = mgh = E down = Iω + ky

61 536 SIMPLE HARMONIC MOTION AND ELASTICITY where the moment of inertia of the bar is I = 3 ml, L = bar length, and ω = v/l. Substituting these into the energy equation, noting from the drawing accompanying the problem statement that ( ) ( ) ( ) h = L, and solving for v, we find that y = 0.00 m m 0.00 m = 0.4 m and that v = 3 ( mgl ky ) m ( )( )( ) ( )( ) kg 9.80 m/s 0.00 m 5.0 N/m 0.4 m = =.08 m/s kg 88. REASONING The change in length of the wire is, According to Equation 0.7, ΔL = FL 0 /YA, where the force F is equal to the tension T in the wire. The tension in the wire can be found by applying Newton's second law to the two crates. SOLUTION The drawing shows the free-body diagrams for the two crates. Taking up as the positive direction, Newton's second law for each of the two crates gives T m g = m a () T m g = m a () T m g T m g Solving Equation () for a, we find a = T m g. Substituting into Equation () gives m Solving for T we find T m g + m m T = 0 T = m m g = (3.0 kg)(5.0 kg)(9.80 m/s ) = 37 N m + m 3.0 kg kg Using the value given in Table 0. for Young s modulus Y of steel, we find, therefore, that the change in length of the wire is given by Equation 0.7 as

62 Chapter 0 Problems 537 (37 N)(.5 m) ΔL = (.0 0 N/m )( m ) =. 0 5 m 89. REASONING If we compare Equation 0.7, which governs the stretching and compression of a solid cylinder, with Equation 0., we find that x is analogous to ΔL and k is analogous to the term YA/L 0 : SOLUTION a. Solving for k we have 0 0 F = YA L 0 ΔL x k ( ) ( 6 πr 3. 0 N/m ) π( m) YA Y k = = = = L L.5 0 m 3. 0 N/m b. The work done by the variable force is equal to the area under the F-versus-x curve. The amount x of stretch is F N x = = = m k 3. 0 N/m The work done is ( )( ) W F x = = = N m.4 0 J 90. REASONING The angular frequency for simple harmonic motion is given by Equation 0. as ω = k/ m. Since the frequency f is related to the angular frequency ω by f = ω/(π) and f is related to the period T by f = /T, the period of the motion is given by T π = = π ω k m SOLUTION a. When m = m = 3.0 kg, we have that 3.0 kg T = T = π = 0.99s 0 N/m Both particles will pass through the position x = 0 m for the first time one-quarter of the way through one cycle, or

63 538 SIMPLE HARMONIC MOTION AND ELASTICITY 0.99 s Δ t = T 4 = T 4 = = s b. T = 0.99 s, as in part (a) above, while 7.0 kg T = π = 3.0s 0 N/m Each particle will pass through the position x = 0 m every odd-quarter of a cycle, 3 5 T, T, T,... Thus, the two particles will pass through x = 0 m when kg particle 3 5 t = T, T, T, K kg particle 3 5 t = T, T, T, K Since T = 3T, we see that both particles will be at x = 0 m simultaneously when t = T,or t = T = T. Thus, ( ) t = T = 0.99 s = 0.75 s 9. REASONING AND SOLUTION The frequency f of the simple harmonic motion is given by Equations 0.6 and 0. as f = ( / π ) k/ m. If we compare Equation 0.7, which governs the stretching and compression of a solid rod with Equation 0., we find that x is analogous to ΔL and k is analogous to the term YA/L 0 : F = YA L ΔL 0 x k The value of Young s modulus for copper is given in Table 0.. Assuming that the rod has a circular cross-section, its area A is equal to π r, and we have f 0 0 ( π r ) k YA Y = = = π m π L m π L m 3 (. 0 N/m ) π ( m) ( )( ) = = 66 Hz π.0 m 9.0 kg

64 Chapter 0 Problems REASONING The figure shows the force F that the jeep exerts on the lower end of the cable, thereby stretching it. We can find the amount ΔL that the cable stretches by using Equation 0.7, ΔL = FL 0 / (YA), where F is the magnitude of the force, L 0 and A are the unstretched length and cross-sectional area of the cable, respectively, and Y is Young's modulus for steel. All the quantities, except F, in this equation are known or can be readily determined. We will employ Newton's second law to find F, since the upward acceleration and mass of the jeep are known. The amount ΔL that the cable stretches is related to the magnitude F of the stretching force by Equation 0.7: F = Y ΔL L 0 Since the cross section of the cable is circular, its area is A = πr, where r is the radius. Solving Equation 0.7 for the change ΔL in the cable's length, and substituting πr for the area, we have A. ΔL = FL 0 YA = FL 0 Y (πr ) (). All the variables on the right side of this equation are known, except F, and we will obtain an expression for it next. To evaluate the magnitude F of the stretching force, we turn to Newton's second law. Figure 0.9a shows that the force exerted on the lower end of the cable by the jeep is F. According to Newton's third law, the action reaction law, the force exerted on the jeep by the cable is + F. Figure 0.9b shows the free-body diagram for the jeep and the two forces that act on the jeep: the force + F pulling it upward and its downward-acting weight W = mg (see Equation 4.5). According to Newton's second law (Equation 4.b), the net force F y acting on the jeep in the y direction is equal to its mass