Question

A 0.600-kg particle has a speed of 2.00 m/s at point A and kinetic energy of 7.50J at point B. What is (a) its kinetic energy at A, (b) its speed at B, and (c) the net work done on the particle by external forces as it moves from A to B?

Answer 1

Its kinetic energy at A is

$$E_{KA}=\frac{mv^2}{2}=1.2\text{ J}.$$

Its speed at B is

$$v=\sqrt{\frac{2E_{KB}}{m}}=5\text{ m/s}.$$

The net work done on the particle by external forces as it moves from A to B is

$$W=\Delta E_K=E_{KB}-E_{KA}=6.3\text{ J}.$$