A hoop of radius `2 m` weight `100 kg`.It rolls along a horizontal floor so that its centre of mass has a speed of `20 cm s^-1`. How much work has to be done to stop it ?

Options

A. 1
B. 20
C. 2
D. 4

Answer

D

Explanation

Given, hoop is rolling (i.e. pure rolling)
So it will have angular velocity, `omega=vr`

`therefore` Total kinetic energ
`K=(1)/(2)mv^(2)+(1)/(2)Iomega^(2)=(1)/(2)mv^(2)+(1)/(2)mr^(2)((v)/(r))^(2)`
`=(1)/(2)mv^(2)+(1)/(2)mv^(2)" "[because I=mr^(2) and omega=(v)/(r)]`
`=mv^(2)=100xx((20)/(100))^(2)=4`