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A loop of radius 3 meter and weighs `150kg`. It rolls along a horizontal floor so that its centre of mass has a speed of `15 cm//sec`. How much work has to be done to stop it -A. `3.375 J`B. `7.375 J`C. `5.375 J`D. `9.375 J` |
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Answer» Correct Answer - A Required work `=` Total K.E. `= 1/2 mv^(2) (1+(k^(2))/(R^(2)))` `= 1/2 Mv^(2) [1+(k^(2))/(R^(2))]` `= 1/2 xx 150 xx (0.15)^(2) (1+1)` `= 3.375 J` |
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