Chapter 10.1, Problem 18SSC

To determine

Work done by mover on the refrigerator.

Answer to Problem 18SSC

  $W=3.46\text{kJ}$

Explanation of Solution

Given:

Mass of the refrigerator is $m=185\text{kg}$ .

Length of ramp is $l=10\text{m}$ .

Angle of ramp is $\theta =1{\text{1}}^{\text{o}}$ .

Formula used:

Work done can be obtained by:

  $W=Fd$

Where, $F$ is force applied, $W$ is work done and $d$ is displacement along force.

  

The component of $mg$ along plane is $mg\sin \theta$ as shown above.

As the plane is frictionless and refrigerator is moved slowly with constant velocity, it means that there is no net force parallel to ramp on refrigerator.

So, force parallel to plane will be

  $F=mg\sin \theta$

So, work done is

  $\begin{array}{c}W=mg\sin \theta \times d\\ =185\times 9.8\times \sin 11°\times 10\\ =3.46\text{kJ}\end{array}$

Conclusion:

Therefore, work done by mover on the refrigerator is $W=3.46\text{kJ}$ .