Question

The free-body diagram will be identical to the one we drew in the example of the frictionless plane, except we will have a vector for the force of friction in the negative X-direction. What is the work done on the box by the force of kinetic friction?

A.$\mu mgh \sin {30°}$
B.$\mu mgh \tan {30°}$
C.$\mu mgh \cot {30°}$
D.$\mu mgh \cos {30°}$

Answer 1

Hint: To find the work done on the box by the force of kinetic friction, use the formula for work done giving the relation between force of friction and displacement of the box in direction that the force is exerted. But, formula for force of friction is given by product of normal force and coefficient of friction. Substitute the values in the formula for work done. This will give the value for work done on the box by the force of kinetic friction.

Complete answer:
Work done by the force of friction is given by,
$W = F . d$ …(1)
Where, F is the force of friction
            d is the displacement of the box in direction that the force is exerted
Force of friction is given by,
$F= \mu. N$
Where, $\mu$ is the coefficient of friction
            N is the normal force
Substituting the value in above equation we get,
$F = \mu. mg\cos {30°}$ …(2)
Displacement of the box in direction that the force is exerted is given by,
$d = \dfrac {h}{\sin{30°}}$ …(3)
Substituting equation. (2) and (3) in equation. (1) we get,
$W = \mu. mg\cos {30°} \times \dfrac {h}{\sin{30°}}$
$\Rightarrow W= \mu mgh \cot{30°}$
Thus, the work done on the box by the force of kinetic friction is $\mu mgh \cot {30°}$.

So, the correct answer is option D i.e. $\mu mgh \cot {30°}$.

Note:
Students must know that there are two types of frictional forces namely Kinetic friction and Static friction. Kinetic friction is the friction that acts between two surfaces in contact that are moving. While static friction is the friction that acts between two surfaces that are not moving. Objects that weigh less exert downward force than heavier objects. Close contacts increase friction between the two objects.