question_answer

A body of mass 1 kg is moving in a vertical. circular path of radius 1 m. The difference between the kinetic energies at its highest and lowest positions is

A)  \[4\sqrt{5}\,\,J\]

B)  20 J   

C)  10 J       

D)  30 J

Correct Answer: B

Solution :

Kinetic energy is the energy possessed by a body of mass m due to its velocity v. \[KE=\frac{1}{2}m{{v}^{2}}\] At the highest point velocity, \[{{v}_{A}}=\sqrt{rg}\] \[\therefore \]  \[KE=\frac{1}{2}mv_{A}^{2}\] At the lowest point velocity, \[{{v}_{B}}=\sqrt{5\,rg}\]                 \[KE=\frac{1}{2}mv_{A}^{2}\] Difference in\[KE=\frac{1}{2}mv_{B}^{2}-v_{A}^{2}=\frac{1}{2}m\,(5rg-rg)\] \[=2mrg=2\times 1\times 1\times 10=20\,J\]