Chapter 7: Work and Kinetic Energy Selected Solutions

Selected Solutions

9. Since the crate moves horizontally, the angle that the rope makes with the horizontal is the same as the angle between the force and the displacement. Therefore, the work done by the tension is

W = (T cosq)d = (125 N)(cos 40.0^o)(5.0 m) = 480 J

19. (a) We can use the work-energy theorem to determine the total work done on the player W_total = DK.

(b) All of the work was done by kinetic friction so we can write

25. The initial kinetic energy of the block is stored as potential energy in the spring. Therefore,

39. The minimum power needed equals the power required to lift 10.0 lb a distance of 2.00 m in one second at constant speed.

51. (a) The puck slows because of work done by friction. We can use the work-energy theorem to relate the reduction in speed to the coefficient of friction. The work done is W = -f_kd = -m_kNd = -m_kmgd, therefore, . Solving this for m_k gives.

(b) Now that we know the coefficient of friction we can solve for the final speed using 44 m/s as the initial speed. This gives

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Selected Solutions by David Reid, Eastern Michigan University. ©2002 by Prentice Hall, Inc.

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