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

(b) Here, `R = 2 m, M = 100 kg, v = 200 cm s^-1`
=`20 xx 10^-2 m s^-1`
Total kinetic of the loop `= K_T + K_R`
=`(1)/(2) Mv^2 + (1)/(2) I omega^2 ["because For a hoop", I = MR^2]`
=`(1)/(2) Mv^2 + (1)/(2) MR^2omega^2`
=`(1)/(2) Mv^2 + (1)/(2) Mv^2`
=`Mv^2`
`"Work required to stop the hoop = Total kinetic energy of the hoop"`.
=`Mv^2 = (100 kg) (20 xx 10^2 ms^-1) = 4 J`.