A child is moving on a skateboard at the speed of 2 ms−1. After a force acts on him, his speed becomes 3 ms−1. The work done by the external force on the child is ___J.
Given, the weight of the child is 20 kg.
Let initial velocity of the child be v1 and final velocity be v2
v1=2 ms−1
v2=3 ms−1
Since the child is in motion, he will have a kinetic energy of mv22, which will be increased after an application of force by an external agent on him.
Work done by the external force on the child =mv222–mv212
=20×(9–4)2
=10×5=50 J