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A body slides down a rough inclined plane and then over a rough plane surface. It starts at A and stops at D. The coefficient of friction is : |
Answer» Solution : `V=sqrt(2G(sin THETA-MU COS theta)AC)` `THEREFORE CD = (V^(2))/(2mu g)` `CD=(cancel(2)cancel(g)(sin theta - mu cos theta)AC)/(cancel(2) mu cancel(g))` `mu CD=((AB)/(AC)-(mu BC)/(AC))AC` `mu CD = AB-mu BCtherefore mu = (AB)/(BD)` |
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