Question

 If a boy lifts a mass of 6kg to a height of 10m and travels horizontally with a constant velocity of 4.2m/s, calculate the work done? Explain your answer.

Answer 1

The work done is zero. In this case the force acts upward on the mass as it is displaced rightward. As a result, the force vector and the displacement vector are at right angles to each other. Therefore, the angle between the force and displacement is $90^{\circ}$ and we get:

$$W=Fdcos\theta=mghcos\theta=6\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot10\ m\cdot cos90^{\circ}=0.$$