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. 2 J
B. 4 J
C. 6 J
D. 8 J

Answer

B

Explanation

Here , `r=2m, M=100kg`,
`v=20 cms^(-1)= 20 xx 10^(-2) ms^(-1)`
Total kinetic energy of the hoop =`K_(T)+K_(R)`
`=1/2 Mv^(2)+1/2Iomega^(2) = 1/2Mv^(2) +1/2MR^(2)omega^(2)`
`=1/2Mv^(2) + 1/2Mv^(2)` `[therefore v=Romega]`
`=Mv^(2)`
Work required to stop the hoop = Total kinetic energy of the hoop
`=Mv^(2)=(100 kg) (20 xx 10^(-2) ms^(-1))^(2) = 4J`