Physics Solution Manual for 1100 and 2101

The magnitude of the displacement or the distance is

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Unformatted text preview: pposes the motion, it is a force that is always directed opposite to the displacement. Therefore, negative work is done. 13. (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. 14. (c) The ball comes to a halt at B, when all of its initial kinetic energy is converted into potential energy. 15. (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. 16. (b) The total mechanical energy is the kinetic energy plus the gravitational potential energy: 2 E = 1 mv 2 + mgh = 1 ( 88.0 kg )(19.0 m/s ) + ( 88.0 kg ) 9.80 m/s 2 ( 55.0 m ) . Since friction 2 2 ( ) 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. 17. (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 KE 0 2.2 J = = 0.45 m . Thus, the conservation principle, KE0 = mgh, or h = mg ( 0.50 kg ) 9.80 m/s 2 ( ) the car can pass over hills A, B, C, and D but not hill E. 18. (a) Using the ground as the zero level for measuring height and applying the energy conservation principle gives the following result: 1 2 m ( 7.00 m/s ) + 1 m ( 9.00 m/s ) + mg (11.0 m ) = 1 mvf2 2 2 2 2 The mass m of the ball can be eliminated algebraically. Chapter 6 Answers to Focus on Concepts Questions 283 19. 7.07 m/s 20. (d) The work done by the net nonconservative force is Wnc = KEf + PEf −KE0 − PE0. 21. (b) The principle of conservation of mechanical energy applies only if the work done by the net nonconservative force is zero, Wnc = 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. 22. (a) A single nonconservative force is the net nonconservative force, and the work it does is given by Wnc = ∆KE + ∆PE. Since the velocity is constant, ∆KE = 0 J. Therefore, Wnc = ∆PE = mg(hf − h0). But hf − h0 = −325 m, since hf is smaller than h0. Thus, Wnc = (92.0 kg)(9.80 m/s2)(−325 m). 23. 71.5 J 24. 12 370 J 25. 344 W 26. 130 N 27. 376 J 284 WORK AND ENERGY CHAPTER 6 WORK AND ENERGY PROBLEMS ______________________________________________________________________________ 1. SSM REASONING The work W done by the tension in the tow rope is given by Equation 6.1 as W = ( F cos θ ) s , where F is the magnitude of the tension, s is the magnitude of the skier’s displacement, and θ is the angle between the tension and the displacement vectors. The magnitude of the displacement (or the distance) is the speed v of the skier multiplied by the time t (see Equation 2.1), or s = v t . SOLUTION Substituting s = v t into the expression for the work, W = ( F cos θ ) s , we have W = ( F cos θ ) v t . Since the skier moves parallel to the boat and since the tow rope is parallel to the water, the angle between the tension and the skier’s displacement is θ = 37.0°. Thus, the work done by the tension is W = ( F cos θ ) vt = [(135 N ) cos 37.0°] ( 9.30 m/s )(12.0 s ) = 1.20 × 104 J ____________________________________________________________________________________________ 2. REASONING a. We will use W = ( F cos θ ) s (Equation 6.1) to determine the work W. The force (magnitude = F) doing the work is the force exerted on you by the elevator floor. This force is the normal force and has a magnitude FN, so F = FN in Equation 6.1. To determine the normal force, we will use the fact that the elevator is moving at a constant velocity and apply Newton’s second law with the acceleration set to zero. Since the force exerted by the elevator and the displacement (magnitude = s) are in the same direction on the upward part of the trip, the angle between them is θ = 0°, with the result that the work done by the force is positive. b. To determine the normal force, we will again use the fact that the elevator is moving at a constant velocity and apply Newton’s second law with the acceleration set to zero. Since the force exerted by the elevator and the displacement are in opposite directions on the downward part of the trip, the angle between them is θ = 180°, and so the work done by the force is negative. SOLUTION a. The fr...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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