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 net work done on an object is equal to the change in its kinetic energy. 
The initial kinetic energy of the cart is (1/2)mv^2 = (1/2)(5.0 kg)(6.0 m/s)^2 = 90 J.
The final kinetic energy of the cart is (1/2)mv^2 = (1/2)(5.0 kg)(10.0 m/s)^2 = 250 J.
Therefore, the change in kinetic energy is: 
250 J - 90 J = 160 J. 
Thus, the net work done on the cart must be 160 J, which is choice D.

A 50-Kg Cart Is Moving Horizontally at 6