A 5.0-kg cart is moving horizontally at 6.0 m/s. In order to change its speed to 10.0 m/s, the net work done on the cart must be:

Question

Quizplus · Accepted Answer

The change in kinetic energy is:$$\Delta K = K_f - K_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$$$\Delta K = \frac{1}{2}(5.0 	ext{ kg})(10.0 	ext{ m/s})^2 - \frac{1}{2}(5.0 	ext{ kg})(6.0 	ext{ m/s})^2$$$$\Delta K = 160 	ext{ J}$$Therefore, the net work done on the cart must be 160 J.

A 50-Kg Cart Is Moving Horizontally at 6