InterviewSolution
Saved Bookmarks
| 1. |
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 ?A. 2 JB. 4 JC. 6 JD. 8 J |
|
Answer» Correct Answer - B (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`. |
|