question_answer

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

A) \[\text{4}\sqrt{5}J\]                      

B) \[20J\]

C) \[10J\]                                 

D) \[30J\]

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, \[{{v}_{B}}=\sqrt{5rg}\] \[KE=\frac{1}{2}mv_{B}^{2}\] Difference in \[KE=\frac{1}{2}m(v_{B}^{2}-v_{A}^{2})\] \[=\frac{1}{2}m(5rg-rg)\] \[=2mrg=2\times 1\times 1\times 10=20J\]