Question

A body of mass m is moving along the circular track of radius r with a constant speed v. The force on the body is $\frac{m{v}^2}{r}$ and is directed towards the centre. The work done by this force in moving the body over half the circumference of the circle is

• A. $\frac{m{v}^2}{r}\times \mathrm{\pi r}$
• B. $\frac{m{v}^2}{{\mathrm{\pi r}}^2}$
• C. $\frac{{\mathrm{\pi r}}^2}{m{v}^2}$
• D. zero

Answer 1

The correct option is D

zero

Step 1: Given data

Force on the body, f = $\frac{m{v}^2}{r}$

The direction of the force is towards the center

The direction of displacement is tangentially along the circumference of the circle.

Step 2: Calculation

Finding the angle $\theta$,

$\theta$ is the angle between the force and direction of displacement,

So here, $\theta$=90^o

The work done is given as,

$$\begin{gathered} Workdone=force\times displacement\times \cos \theta \\ =force\times displacement\times \times \cos 90 \\ =0(\sin ce\cos 90=0) \end{gathered}$$

Hence, option (D) is the correct option.