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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, `= 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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