A batter hits a baseball with mass 0.145 kg straight upward with an initial speed of 25.0 m/s. (a) How much work has gravity done on the baseball when it reaches a height of 20.0 m above the bat? (b) Use the work-energy theorem to calculate the speed of the baseball at a height of 20.0 m above the bat. You can ignore air resistance. (c) Does the answer to part (b) depend on whether the baseball is moving upward or downward at a height of 20.0 m? Explain.
Solution 26E Introduction We have to calculate the work done using the force and displacement. Then we have to calculate the final kinetic energy and from that we can calculate the velocity of the ball at that height. Step 1 The gravitational force is given by w = mg = (0.145 kg)(9.8 m/s ) = 14.21 N This force will act vertically downwards and the displacement is vertically upwards. Hence the angle between the gravitational force and the displacement is 180°. So the work done by gravity is W = wdcos = (14.12 N)(2.00 m)cos(180°) = 28.24 J -28.24 J of work has been done by gravity on the ball. Step 2 The initial kinetic energy of the ball is And the work done, as calculated above, is W = 28.24 J. So the total energy kinetic energy of the ball at that height is Hence, if the velocity at that point2is v , then we have Hence the velocity at 2.0 m height will be 15.3 m/s.