Browse Questions

A man cleaning a floor pulls a vacuum cleaner with a force of magnitude F = 50.0 N at an angle of $30.0^{\large\circ}$ with the horizontal. Calculate the work done by the force on the vacuum cleaner as the vacuum cleaner is displaced 3 m to the right.

$\begin{array}{1 1}130J\\150J\\120J\\100J\end{array}$

The work done by the force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus W = $(F \cos \theta )d = F.d$
Using the definition of work
W = $50 \times \cos 30 \times 3$
$\;\;\;=130N.m$
$\;\;\;=130J$