Suppose the sled in Exercise 6.36 is initially at rest at x = 0. Use the work–energy theorem to find the speed of the sled at (a) x = 8.0 m and (b) x = 12.0 m. Ignore friction between the sled and the surface of the pond. 6.36 . A child applies a force parallel to the x -axis to a 10.0-kg sled moving on the frozen surface of a small pond. As the child controls the speed of the sled, the x -component of the force she applies varies with the x -coordinate of the sled as shown in ?Fig. E6.36.? Calculate the work done by when the sled moves (a) from x = 0 to x = 8.0 m; (b) from x = 8.0 m to x = 12.0 m; (c) from x = 0 to 12.0 m.

Solution 35E Step 1: Step 2: Consider work done W = F * x W = Area under F v/s x graph. W = K.E All work being converted to kinetic energy because there is no friction or potential energy does not change. Speed of the sled at x = 8.0 m. 2 ½ mv = 40 J ½ (10 kg)v = 40 J 5v = 40 J v = 2.82 m/s. Speed of the sled at x = 12.0 m. ½ mv = 60 J ½ (10 kg)v = 60 J 5v = 60 J v = 3.46 m/s Step3: (a).work done by F when sleds move from x = 0 m to x = 8.0 m. W =½ F (N)x x W =½ (10 N)(8 m) W = 40 J.