POTENTIAL ENERGY AND ENERGY CONSERVATION

Transcription

1 7 POTENTIAL ENERGY AND ENERGY CONSERVATION 7.. IDENTIFY: U grav = mgy so ΔU grav = mg( y y ) SET UP: + y is upward. EXECUTE: (a) ΔU = (75 kg)(9.8 m/s )(4 m 5 m) = J (b) ΔU = (75 kg)(9.8 m/s )(35 m 4 m) = J EVALUATE: U grav increases when the altitude of the object increases IDENTIFY and SET UP: Use energy methods. (a) K + U + U. Solve for K and then use K = mv to obtain v. Figure 7.5 = (The only force on the ball while it is in the air is gravity.) K = mv ; K = mv U = mgy, y =. m U = mgy =, since y = for our choice of coordinates. EXECUTE: mv + mgy = mv v = v + gy = (. m/s) + (9.8 m/s )(. m) = 4. m/s EVALUATE: The projection angle of 53. doesn t enter into the calculation. The kinetic energy depends only on the magnitude of the velocity; it is independent of the direction of the velocity. (b) Nothing changes in the calculation. The expression derived in part (a) for v is independent of the angle, so v = 4. m/s, the same as in part (a). (c) The ball travels a shorter distance in part (b), so in that case air resistance will have less effect. 7.. IDENTIFY: Apply Eq.(7.7) to the motion of the car. SET UP: Take y = at point A. Let point be A and point be B. K + U + U EXECUTE: U =, U = mg(r) = 8,4 J, = W f K = mv = 37,5 J, K = mv = 384 J The work-energy relation then gives W f + U K = 54 J.

2 EVALUATE: Friction does negative work. The final mechanical energy (K + U = 3,64 J) is less than the initial mechanical energy (K + U = 37,5 J) because of the energy removed by friction work IDENTIFY: Only gravity does work, so apply Eq.(7.4). Use F = ma to calculate the tension. SET UP: Let y = at the bottom of the arc. Let point be when the string makes a 45 angle with the vertical and point be where the string is vertical. The rock moves in an arc of a circle, so it has radial acceleration a rad = v / r EXECUTE: (a) At the top of the swing, when the kinetic energy is zero, the potential energy (with respect to the bottom of the circular arc) is mgl( cos ), where l is the length of the string and is the angle the string makes with the vertical. At the bottom of the swing, this potential energy has become kinetic energy, so mgl( cos ) = mv, or v = gl( cos ) = (9.8 m/s ) (.8 m) ( cos45 ) =. m/s. (b) At 45 from the vertical, the speed is zero, and there is no radial acceleration; the tension is equal to the radial component of the weight, or mg cosθ = (. kg) (9.8 m/s ) cos 45 =.83 N. (c) At the bottom of the circle, the tension is the sum of the weight and the mass times the radial acceleration, mg + mv l = mg(+ ( cos45 )) =.9 N EVALUATE: When the string passes through the vertical, the tension is greater than the weight because the acceleration is upward IDENTIFY: Use the information given in the problem with F = kx to find k. Then U el = kx. SET UP: x is the amount the spring is stretched. When the weight is hung from the spring, F = mg. EXECUTE: U el k = F x = mg x = (3.5 kg)(9.8 m/s ) = 5 N/m..34 m. m (. J) x = ± = ± = ±.95 m = ±9.5 cm. The spring could be either stretched 9.5 cm k 5 N/m or compressed 9.5 cm. If it were stretched, the total length of the spring would be. cm cm =.5 cm. If it were compressed, the total length of the spring would be. cm 9.5 cm =.48 cm. EVALUATE: To stretch or compress the spring 9.5 cm requires a force F = kx = N IDENTIFY and SET UP: Use energy methods. There are changes in both elastic and gravitational potential energy; elastic; U = kx, gravitational: U = mgy. EXECUTE: (a) U = kx so x = U k = (3. J) =.63 m = 6.3 cm 6 N/m (b) Points and in the motion are sketched in Figure 7.9. K + U + U Figure 7.9 = (Only work is that done by gravity and spring force) K =, K = y = at final position of book U = mg ( h + d), U = kd

3 + mg(h + d) + = kd The original gravitational potential energy of the system is converted into potential energy of the compressed spring. kd mgd mgh = d = k mg ± (mg) + 4 k (mgh) d must be positive, so d = k ( mg + (mg) + kmgh) d = 6 N/m ((. kg)(9.8 m/s ) + d =.74 m +.87 m =. m = cm ((. kg)(9.8 m/s )) + (6 N/m)(. kg)(9.8 m/s )(.8 m) EVALUATE: It was important to recognize that the total displacement was h + d; gravity continues to do work as the book moves against the spring. Also note that with the spring compressed. m it exerts an upward force (9 N) greater than the weight of the book (.8 N). The book will be accelerated upward from this position IDENTIFY: Apply Eq.(7.3) and F = ma. SET UP: =. There is no change in U grav. K =, U =. EXECUTE: kx = mv. The relations for m, v x x, k and x are kx = mv x and kx = 5mg. Dividing the first equation by the second gives x = v x, and substituting this into the second gives 5g k = 5 mg. v x (a) k = 5 (6 kg)(9.8 m/s ) (.5 m/s) = N/m (b) x = (.5 m/s) 5(9.8 m/s ) =.8 m EVALUATE: Our results for k and x do give the required values for a x and v x : a x = kx m = ( N/m)(.8 m) 6 kg = 49. m/s = 5.g and v x = x k =.5 m/s. m 7.8. IDENTIFY and SET UP: The force is not constant so we must use Eq.(6.4) to calculate W. The properties of work done by a conservative force are described in Section 7.3. W = F dl, F = α x î EXECUTE: (a) dl = dyĵ (x is constant; the displacement is in the + y-direction ) F dl = (since î ĵ = ) and thus W =. (b) d l = î F d l = ( α x î ) (î ) = α x x W = ( α x ) = x 3 ax3 x = α (x 3 x 3 N/m x 3 ) = ((.3 m) 3 (. m) 3 ) =. J 3

4 (c) d l = î as in part (b), but now x =.3 m and x =. m W = α(x 3 x 3 3 ) = +. J (d) EVALUATE: The total work for the displacement along the x-axis from. m to.3 m and then back to. m is the sum of the results of parts (b) and (c), which is zero. The total work is zero when the starting and ending points are the same, so the force is conservative. EXECUTE: W x = α(x 3 x 3 x 3 ) = α x 3 α x The definition of the potential energy function is W x x = U U. Comparison of the two expressions for W gives U = 3 α x3. This does correspond to U = when x =. EVALUATE: In part (a) the work done is zero because the force and displacement are perpendicular. In part (b) the force is directed opposite to the displacement and the work done is negative. In part (c) the force and displacement are in the same direction and the work done is positive IDENTIFY and SET UP: Use Eq.(7.7) to calculate the force from U (x). Use coordinates where the origin is at one atom. The other atom then has coordinate x. EXECUTE: F x = du = d C 6 x 6 = +C d 6 x 6 = 6C 6 x 7 The minus sign mean that F x is directed in the x-direction, toward the origin. The force has magnitude 6C 6 / x 7 and is attractive. EVALUATE: U depends only on x so F is along the x-axis; it has no y or z components. du (x) IDENTIFY: Apply F(x) =. d( / x) SET UP: = x EXECUTE: F x (x) = d( Gm m / x) x-direction. This is toward m, so the force is attractive. d( / x) = Gm m = Gm m. The force on m x is in the EVALUATE: By Newton's 3 rd law the force on m due to m is Gm m / x, in the +x-direction (toward m ). The gravitational potential energy belongs to the system of the two masses IDENTIFY: Apply Eq.(7.6). du SET UP: is the slope of the U versus x graph. EXECUTE: (a) Considering only forces in the x-direction, F x = du and so the force is zero when the slope of the U vs x graph is zero, at points b and d. (b) Point b is at a potential minimum; to move it away from b would require an input of energy, so this point is stable. (c) Moving away from point d involves a decrease of potential energy, hence an increase in kinetic energy, and the marble tends to move further away, and so d is an unstable point. EVALUATE: At point b, F x is negative when the marble is displaced slightly to the right and F x is positive when the marble is displaced slightly to the left, the force is a restoring force, and the equilibrium is stable. At point d, a small displacement in either direction produces a force directed away from d and the equilibrium is unstable.

5 7.47. (a) IDENTIFY: Use work-energy relation to find the kinetic energy of the wood as it enters the rough bottom. SET UP: Let point be where the piece of wood is released and point be just before it enters the rough bottom. Let y = be at point. EXECUTE: U gives K = mgy = 78.4 J. IDENTIFY: Now apply work-energy relation to the motion along the rough bottom. SET UP: Let point be where it enters the rough bottom and point be where it stops. K + U + U EXECUTE: = W f = µ k mgs, K = U = U = ; K = 78.4 J 78.4 J µ k mgs = ; solving for s gives s =. m. The wood stops after traveling. m along the rough bottom. (b) Friction does 78.4 J of work. EVALUATE: The piece of wood stops before it makes one trip across the rough bottom. The final mechanical energy is zero. The negative friction work takes away all the mechanical energy initially in the system IDENTIFY: Use the work-energy theorem, Eq.(7.7). Solve for K and then for v. SET UP: Let point be at his initial position against the compressed spring and let point be at the end of the barrel, as shown in Figure Use F = kx to find the amount the spring is initially compressed by the 44 N force. K + U + U Figure 7.53 Take y = at his initial position. EXECUTE: K =, K = mv = W fric = fs = (4 N)(4. m) = 6 J U,grav =, U,el = kd, where d is the distance the spring is initially compressed. F = kd so d = F k = 44 N N/m = 4. m and U,el = ( N/m)(4. m) = 88 J U,grav = mgy = (6 kg)(9.8 m/s )(.5 m) = 47 J, U,el = Then K + U + U gives 88 J 6 J = mv + 47 J mv = 77 J and v = (77 J) 6 kg = 5.5 m/s EVALUATE: Some of the potential energy stored in the compressed spring is taken away by the work done by friction. The rest goes partly into gravitational potential energy and partly into kinetic energy IDENTIFY: Conservation of energy says the decrease in potential energy equals the gain in kinetic energy. SET UP: Since the two animals are equidistant from the axis, they each have the same speed v.

6 EXECUTE: One mass rises while the other falls, so the net loss of potential energy is (.5 kg. kg)(9.8 m/s )(.4 m) =.76 J. This is the sum of the kinetic energies of the animals and is equal to m tot v, and v = (.76 J) =.83 m/s. (.7 kg) EVALUATE: The mouse gains both gravitational potential energy and kinetic energy. The rat s gain in kinetic energy is less than its decrease of potential energy, and the energy difference is transferred to the mouse IDENTIFY: Apply Eq.(7.4) to the motion of the package. = W fk, the work done by the kinetic friction force. SET UP: f k = µ k n = µ k mg cosθ, with θ = 53.. Let L = 4. m, the distance the package moves before reaching the spring and let d be the maximum compression of the spring. Let point be the initial position of the package, point be just as it contacts the spring, point 3 be at the maximum compression of the spring, and point 4 be the final position of the package after it rebounds. EXECUTE: (a) K =, U =, = f k L = µ k Lcosθ. U = mglsinθ. K = mv, where v is the speed before the block hits the spring. Eq.(7.4) applied to points and, with y =, gives U. Solving for v, v = gl(sinθ µ k cosθ) = (9.8 m/s )(4. m)(sin53. (.)cos53. ) = 7.3 m/s. (b) Apply Eq.(7.4) to points and 3. Let y 3 =. K = K 3 =. U = mg(l + d)sinθ. U = kd. = f k (L + d). Eq.(7.4) gives mg(l + d)sinθ µ k mg cosθ(l + d) = kd. This can be written as d k d L =. The factor multiplying d is mg(sinθ µ k cosθ) 4.54 m, and use of the quadratic formula gives d =.6 m. (c) The easy thing to do here is to recognize that the presence of the spring determines d, but at the end of the motion the spring has no potential energy, and the distance below the starting point is determined solely by how much energy has been lost to friction. If the block ends up a distance y below the starting point, then the block has moved a distance L + d down the incline and L + d y up the incline. The magnitude of the friction force is the same in both directions, µ k mg cos, and so the work done by friction is µ k (L + d y)mg cos. This must be equal to the change in gravitational potential energy, which is mgysin. Equating these and solving for y gives µ y = (L + d) k cosθ µ = (L + d) k. Using the value of d found in part (b) and the given sinθ + µ k cosθ tanθ + µ k values for µ k and θ gives y =.3 m. EVALUATE: Our expression for y gives the reasonable results that y = when µ k = ; in the absence of friction the package returns to its starting point IDENTIFY: a x = d x/dt, a y = d y/dt. F x = ma x, F y = ma y. U = F x + F y dy. SET UP: d dt (cos t) = sin t. d dt (sin t) = cos t. sin t dt = cosω t. v x = / dt, v y = dy / dt. E = K + U. EXECUTE: (a) a x = d x/dt = x, F x = ma x = m x. a y = d y/dt = y, F y = ma y = m y cosω t dt = sinω t,

7 (b) U = F x + F y dy = mω x + ydy = mω (x + y ) (c) v x = /dt = x sin t = x ( y/y ). v y = dy/dt = + y cos t = + y (x/x ). (i) When x = x and y =, v x = and v y = y, K = m(v + v x y ) = my ω, U = ω mx and E = K + U = mω (x + y ) (ii) When x = and y = y, v x = x and v y =, K = ω mx, U = mω y and E = K + U = mω (x + y ) EVALUATE: The total energy is the same at the two points in part (c); the total energy of the system is constant.