A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by Paul has magnitude 48.0 N and direction 61.0∘ south of west. How much work does Paul’s force do during a displacement of the crate that is 12.0 m in the direction 22.0∘ east of north?

-570.4 J

Explanation:

Parameters given:

Force = 48N, 61° SW

Displacement = 12m, 22.0 NE

Work done is given by:

W = F*d*cosθ

To find θ:

θ = 61 + 90 + 22 = 172°

=> W = 48 * 12 * cos172

W = - 570.4 J

Hence, the work to be done is - 570.4J.

This value is negative because the applied force is opposite the direction of the displacement.