CHAPTER 6 WORK AND ENERGY
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1 CHAPTER 6 WORK AND ENERGY ANSWERS TO FOCUS ON CONCEPTS QUESTIONS (e) When the force is perpendicular to the displacement, as in C, there is no work When the force points in the same direction as the displacement, as in B, the maximum work is done When the force points at an angle with respect to the displacement but has a component in the direction of the displacement, as in A, the work has a value between zero and the maximum work (b) Work is positive when the force has a component in the direction of the displacement The force shown has a component along the x and along the +y axis Therefore, displacements in these two directions involve positive work 3 (c) The work is given by W = (F cos θ)s, which is zero when F = 0 N, s = 0 m, or θ = 90º 4 78 kg m /s 5 (a) The kinetic energy is KE mv Pythagorean theorem indicates that 6 5 m/s Since the velocity components are perpendicular, the v v v Therefore, East North East North m v v KE 300 kg 500 m/s 800 m/s 7 (e) The work-energy theorem states that W = KE f KE 0 Since work is done, the kinetic energy changes Since kinetic energy is KE mv, the speed v must also change Since the instantaneous speed is the magnitude of the instantaneous velocity, the velocity must also change 8 (d) The work-energy theorem indicates that when the net force acting on the particle does negative work, the kinetic energy decreases Since each force does negative work, the work done by the net force must be negative, and the kinetic energy must decrease But when the kinetic energy decreases, the speed must also decrease, since the kinetic energy is proportional to the square of the speed Since it is stated that the speed increases, this answer is not possible 9 (b) The work-energy theorem states that the net work done on the particle is equal to the change in the particle s kinetic energy However, the speed does not change Therefore, the kinetic energy does not change, because kinetic energy is proportional to the square of the
2 speed According to the work-energy theorem, the net work is zero, which will be the case if W = W 0 (d) Since the block starts from rest, the work energy theorem is W mv mv 70 kg v f 0 f The final speed can be obtained from this expression, since the work done by the net force is W = (750 N) (cos 380º) (650 m) + (540 N) (cos 00º) (650 m) + (930 N) (cos 650º) (650 m) (c) The gravitational force is a conservative force, and a conservative force does no net work on an object moving around a closed path (e) Since air resistance always opposes the motion, it is a force that is always directed opposite to the displacement Therefore, negative work is done 3 (d) The velocity is constant Therefore, the speed is also constant, and so is the kinetic energy However, the total mechanical energy is the kinetic plus the gravitational potential energy, and the car moves up the hill, gaining potential energy as it goes Thus, in the circumstance described mechanical energy cannot be conserved 4 (c) The ball comes to a halt at B, when all of its initial kinetic energy is converted into potential energy 5 (a) The stone s motion is an example of projectile motion, in which the final vertical height is the same as the initial vertical height Because friction and air resistance are being ignored, mechanical energy is conserved Since the initial and final vertical heights are the same, PE = 0 J Since energy is conserved, it follows that KE = 0 J also 6 (b) The total mechanical energy is the kinetic energy plus the gravitational potential energy: E mv mgh 880 kg 90 m/s 880 kg 980 m/s 550 m Since friction and air resistance are being ignored, the total mechanical energy is conserved, which means that it has the same value, no matter what the height above sea level is 7 (d) Since the track is frictionless, the conservation of mechanical energy applies We take the initial vertical level of the car as the zero level for gravitational potential energy The highest vertical level that the car can attain occurs when the final speed of the car is zero and all of the initial kinetic energy is converted into gravitational potential energy According to KE0 J the conservation principle, KE 0 = mgh, or h 045 m Thus, mg 050 kg 980 m/s the car can pass over hills A, B, C, and D but not hill E 8 (a) Using the ground as the zero level for measuring height and applying the energy conservation principle gives the following result:
3 700 m/s 900 m/s 0 m m m mg mv f The mass m of the ball can be eliminated algebraically m/s 0 (d) The work done by the net nonconservative force is W nc = KE f + PE f KE 0 PE 0 (b) The principle of conservation of mechanical energy applies only if the work done by the net nonconservative force is zero, W nc = 0 J When only a single nonconservative force is present, it is the net nonconservative force, and a force that is perpendicular to the displacement does no work (a) A single nonconservative force is the net nonconservative force, and the work it does is given by W nc = KE + PE Since the velocity is constant, KE = 0 J Therefore, W nc = PE = mg(h f h 0 ) But h f h 0 = 35 m, since h f is smaller than h 0 Thus, W nc = (90 kg)(980 m/s )( 35 m) 3 75 J J W 6 30 N J 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
4 m 3 (b) According to Equations 04 and 0, the period T is given by T Greater k 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 08 v A max Thus, increases in both the amplitude A and the angular frequency lead to an increase in the maximum speed 5 (e) The maximum acceleration in simple harmonic motion is given by Equation 00 a A A decrease in the amplitude A decreases the maximum acceleration, but this max decrease is more than offset by the increase in the angular frequency, 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 (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 03), where A is the amplitude of the previous simple harmonic motion This total mechanical energy is
5 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 is given by Equation 0 as Thus, when m the mass m attached to the spring decreases, the angular frequency increases 4 (b) The angular frequency of oscillation of a simple pendulum is given by Equation 06 g L It depends only on the magnitude g of the acceleration due to gravity and the length L of the pendulum It does not depend on the mass Therefore, the pendulum with the greatest length has the smallest frequency 5 7 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 m Since the frequency of the force is doubled, the new frequency must be k m 8 The frequency of system A is, in fact, k k m m L 8 (c) According to Equation 07 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 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
ELASTICITY. values for the mass m and smaller values for the spring constant k lead to greater values for the period.
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
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