Question

A force $\left( {\dfrac{{m{v^2}}}{r}} \right)$ is acting on a body of mass $m$ moving with a speed $v$ in a circle of radius $r$ . What is the work done by the force in moving the body over half the circumference of the circle?
A. $\dfrac{{m{v^2}}}{r} \times \pi r$

B. Zero

C. $\dfrac{{m{v^2}}}{{{r^2}}}$

D. $\dfrac{{\pi {r^2}}}{{m{v^2}}}$

Answer 1

Hint: Here we have to apply the concept of centripetal force since the body is moving in a circle.
Centripetal force is defined as the force that is required to hold an object going in a curved path and that is guided inward toward the centre of rotation.

Complete step by step answer:
Given, Force $ = \dfrac{{m{v^2}}}{r}$
Where $m$ is the mass of the body, $v$ is the speed of the body and $r$ is the radius of the body.
The portion of the force which is perpendicular to the velocity is the part which results in the centripetal force.
A centripetal force is a net force acting on an object to keep it travelling along a circular road. It is important to note that the centripetal force is not a fundamental force, but merely a name provided to the net force that induces the object to travel along a circular direction.
Just the direction of the velocity varies in a uniform circular motion, since the force is at the right angle to the movement. No work is performed as the speed is constant and thus the energy remains constant.
So the centripetal force is a circular motion, so the work done by centripetal force is always zero. As the force is working in the centre, the work performed by this force in pushing the body over half the diameter of the circle is zero.

Hence, option B is the correct answer.

Note:The centripetal force is zero as speed remains constant during a circular motion. But without a centripetal force a body cannot move in circular motion.