Question

A 13-N force moves through a displacement of 42 m, at an angle of 35o to the
displacement. What work is now done by the force? What if the angle is 90 degrees?

Answer 1

By definition, the work is:

$$A = \mathbf{F}\cdot \mathbf{s}$$

Where $\mathbf{F}$ is vector of the foce and $\mathbf{s}$ is vector of the displacement. $\cdot$ denotes scalar product. Thus, obtain:

$$A = Fs\cos\theta$$

where $F = 13N$ is the magnitude of the force, $s = 42m$ is the magnitude of the displacement, and $\theta=35\degree$ is the angle between the work and displacement vectors. Get:

$$A = Fs\cos\theta = 13\cdot 42\cdot \cos35\degree \approx 447.3J$$

If $\theta =90\degree$, then $\cos\theta = 0$ and the work is equal to 0 J.

Answer. a) 447.3 J, b) 0 J.