Problem 6-3. You are moving into an apartment. Your weight is 685 N and that of your belongings is 915 N. (a) How much work does the elevator do in lifting you and your belongings up five stories (15.2 m) at a constant velocity? (b) How much work does the elevator do on you (without belongings) on the downward trip, which is also made at constant velocity? Be sure to include the correct sign for the work.

(a) The elevator must exert a force equal to the weight of you and your belongings to lift both at constant velocity. The work done by the elevator is

(b) The force and the displacement are in opposite directions on the downward trip (but the formula automatically takes care of the sign)

Problem 6-15. A 65.0-kg jogger is running at a speed of 5.30 m/s. (a) What is the kinetic energy of the jogger? (b) How much work is done by the net external force that accelerates the jogger to 5.30 m/s from rest?

(a) The kinetic energy is

(b) The work-energy theorem states that the work done equals the increase in the kinetic so, since the initial kinetic energy is zero, the work done is also 913 J.

Problem 6-32. A 75.0-kg skier rides a 2830-m-long lift to the top of a mountain. The lift makes an angle of 14.6^o with the horizontal. What is the change in the skier's gravitational potential energy?

The vertical height of the mountain is

The change in the skier's gravitational potential energy is mgh, which is

Problem 6-52. A pitcher throws a 0.140-kg baseball, and it approaches the bat at a speed of 40.0 m/s. The bat does W_nc = 70.0 J of work on the ball in hitting it. Ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is 25.0 m above the point of impact.

The work done by the non-conservative force equals the increase in the kinetic energy plus the increase in the potential energy

W_nc is 70J, ½mv₀² is 112J, and the increase in potential energy is 34.3 J. So the final kinetic energy is

70J + 112J - 34J = 148J.

From this, the final speed can be found to be 46.0 m/s.

Problem 6-60. In an action-adventure movie, the hero lifts the 91-kg villain straight upward through a distance of 1.2 m in 0.51 s at a constant speed. What power does the hero produce while doing this?

The upward force that the hero must exert is equal in magnitude to the weight mg of the villain. This is 91*9.8 = 892 N. The hero lifts the villain at a speed of 1.2m/0.51s = 2.35 m/s. The power is P = Fv = 2100 W.