Question

A particle of mass m moves in a circular path radius r under the action of a force $\frac{m{v}^2}{r}$ . The work done during its motion over half of the circumference of the circular path will be :

• A. $\left(\frac{m{v}^2}{r}\right)\times 2\pi r$
• B. $\left(\frac{m{v}^2}{r}\right)\times \pi r$
• C. $\left(\frac{2\pi r}{\frac{m{v}^2}{r}}\right)$
• D. Zero

Answer 1

The correct option is D Zero

Work done by centripetal is always zero.