Question

A small block of mass $m$ is kept on a rough inclined surface of inclination $\theta$ fixed in an elevator. The elevator goes up with a uniform velocity $v$ and the block does not slide on the wedge. The work done by the force of friction on the block in time $t$ will be
(A) Zero
(B) $- mgvt{\cos ^2}\theta$
(C) $- mgvt{\sin ^2}\theta$
(D) $mgvt\sin 2\theta$

Answer 1

Friction is the force that resists motion. Work done is the product of force exerted and the displacement, as the result of force.
Formula Used: The formulae used in the solution are given here.
Work done is thus given by, $W = F \cdot d\cos \left( {{{90}^ \circ } - \theta } \right)$ where $F$ is the force applied and $d$ is the displacement and $\left( {{{90}^ \circ } - \theta } \right)$ is the angle between the two.

Complete Step by Step Solution: It has been given that a small block of mass $m$ is kept on a rough inclined surface of inclination $\theta$ fixed in an elevator. The elevator goes up with a uniform velocity $v$ and the block does not slide on the wedge.
The vertical distance travelled by the elevator in time $t$when travelling with velocity $v$ is = $vt$.
Friction is the force that resists motion when the surface of one object comes in contact with the surface of another. In other words, frictional force refers to the force generated by two surfaces that contact and slide against each other.
As the block does not slide on the wedge. So force of friction is given by $F = \mu N$ where $\mu$ is the coefficient of friction and $N$ is the normal force.
The vertical component of the force of friction is, $F = mg\sin \theta$.
Work done is the product of force exerted and the displacement, as the result of force.
The angle between $F$ and $d$ is measured to be $\left( {{{90}^ \circ } - \theta } \right)$.
Work done is thus given by, $W = F \cdot d\cos \left( {{{90}^ \circ } - \theta } \right)$ where $F$ is the force applied and $d$ is the displacement and $\left( {{{90}^ \circ } - \theta } \right)$ is the angle between the two.
Since, $F = mg\sin \theta$ and $d = vt$, we assign these values in the equation for work.
Therefore work done is, $W = mg\sin \theta \cdot vt \cdot \cos \left( {90 - \theta } \right)$
Multiplying the values, we get, $W = mgvt{\sin ^2}\theta$.
But the work done by the force of friction on the block in time $t$ will be $W = - mgvt{\sin ^2}\theta$ since the force of friction is in the opposite direction that the velocity.

The correct answer is Option C.

Note: The forces of friction are mainly affected by the surface texture and amount of force impelling them together and the angle and position of the object affect the amount of frictional force.
If an object is placed flat against an object, then the frictional force will be equal to the weight of the object. If an object is pushed against the surface, then the frictional force will be increased and becomes more than the weight of the object.