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. $\frac{(2 \pi r)}{\left(\frac{m v^{2}}{r}\right)}$
• D. zero

Answer 1

Answer: (D)

Work done by centripetal force is always zero.