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A body of mass 'm' slides down a smooth inclined plane having an inclination of 45^(@)with the horizontal. It takes 2S to reach the bottom. It the body is placed on a similar plane having coefficient friction 0.5 What is the time taken for it to reach the bottom ? |
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Answer» Solution :Mass = m, `theta = 45^(@)` Time taken to reach the bottom `T_(1) =2 SEC mu = 0.5` Time taken by the body to touch the bottom without friction is `T_(1) = sqrt((2L)/(g SIN theta))` Time taken with friction is `T_(2) = sqrt((2l)/(g(sin theta- mu.cos theta)))` `T_(1)/T_(2) = sqrt((sin theta. mu cos theta)/(sin theta))` `T_(2) = T_(1) sqrt((sin theta)/(sin theta- mu cos theta)) = 2.828` sec |
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