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An object is placed on the surface of a smooth inclined plane of inclination theta. It takes time t to reachthe bottom. If the same object is allowed to slide down a rough inclined plane of inclination theta, it takes time nt to reach the bottom where n is a number greater than 1. The coefficient of friction mu is given by:- |
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Answer» `mu=tan theta(1-1/n^2)` If s be the distance travelled , then `s=1/2 g sin theta xx t_1^2` ….(i) On rough inclined plane : Acceleration , `a=(mg sin theta -muR)/m` or `a=(mg sin theta -mu mg cos theta )/m` `=g sin theta - mug cos theta` `therefore s=1/2 (g sin theta - mug cos theta )t_2^2`...(ii) From equations(i) and (ii) , `t_2^2/t_1^2=(sin theta)/(sin theta - mu cos theta )` But `t_2=nt_1 , therefore n^2=(sin theta)/(sin theta - mu cos theta)` or `mu=(n^2-1)/n^2 xx (sin theta)/(cos theta)` or `mu=(1-1/n^2) tan theta` |
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