Question

A 5.0-kg object is pulled along a horizontal surface at a constant speed by a 15-n force acting 20° above the horizontal. How much work is done by this force as the object moves 6.0 m?

Answer 1

Answer:

84.6 J

Explanation:

The work done by the force is given by

$W=Fd cos \theta$

where

W is the work done

F = 15 N is the force applied

d = 6.0 m is the displacement

$\theta=20^{\circ}$ is the angle between the force's direction and the displacement

Substituting the numbers into the equation, we find

$W=(15 N)(6.0 m)cos 20^{\circ} =84.6 J$