Answer to Question #158703 in Physics for Annabel

Question

An object (m=5kg) moves from 15 m/s to 10 m/s in 5 sec. what is its change in kinetic energy?

Answer 1

$$\Delta KE=\dfrac{1}{2}m(v_f^2-v_i^2),$$ $$\Delta KE=\dfrac{1}{2}\cdot 5\ kg\cdot((10\ \dfrac{m}{s})^2-(15\ \dfrac{m}{s})^2)=-312.5\ J.$$

The sign minus means that there is a negative work done upon object to decelerate it.