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 position is :

A)  20 J         

B)  10 J

C)  \[4\sqrt{5}J\]       

D)  10\[(\sqrt{5}-1)J\]

Correct Answer: A

Solution :

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