Question

A box is pulled across a level floor a distance of 0.1 km. Given that 2.0 kJ of work was done in overcoming friction, what was the average friction force?

Answer 1

$$W=Fd\cos\theta$$

$$F=\frac{W}{d}=\frac{2000\: \rm J}{100\: \rm m}=20\:\rm N$$