A solid sphere of mass m rolls down an inclined plane without slipping from rest at the top of an inclined plane.The linear speed of the sphere at the bottom of the inclined plane is v .The kinetic energy of the sphere at the bottom is

Options

A. `(1)/(2)mv^(2)`
B. `(5)/(3)mv^(2)`
C. `(2)/(5)mv^(2)`
D. `(7)/(10)mv^(2)`

Answer

D

Explanation

The kinetic energy of the solid sphere at the bottom is
`K=K_(T)+K_(R)=(1)/(2)mv^(2)+(1)/(2)lomega^(2)`
`=(1)/(2)mv^(2)+(1)/(5)((2)/(5)mR^(2))omega^(2)`
[for solid sphere,`I=(2)/(5)mR(2)]`
`=(1)/(2)mv^(2)+(1)/(5)mv^(2)` [As v=`Romega`]
`=mv^(2)[(1)/(2)+(1)/(5)]=(7)/(10)mv^(2)`