Question

A uniform thin ring of mass 0.4 kg rolls without slipping on a horizontal surface with a linear velocity of 10 cm/s. The kinetic energy of the ring is

• A. $ 4\times {{10}^{-3}} $
• B. $ 4\times {{10}^{-2}}J $
• C. $ 2\times {{10}^{-3}}J $
• D. $ 2\times {{10}^{-2}}J $

Answer 1

Answer: (A)

Given, $ m=0.4\,\,kg $ and $ v=10\,\,cm/s $ Kinetic energy $ =\frac{1}{2}m{{v}^{2}}(1+1) $ Kinetic energy $ =m{{v}^{2}} $ $ =0.4\times {{(0.1)}^{2}} $ $ =4\times {{10}^{-3}}J $