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A body is released from rest from the top of an inclined plane of length 'L' and angle of inclination 'theta'. The top of plane of length (1)/(n)L(n gt 1) is smooth and the remaining part is rough. If the body comes to rest on reaching the bottom of the plane then find the value of coefficient of friction of rough plane. |
Answer» Solution : For smooth part : `V^(2)=2a_(1)(L)/(n)` For rough part : `0-V^(2)=2a_(2)((n-1)/(n))L` `2a_(1)(L)/(n)=-2a_(2)((n-1)/(n))L` `g SIN THETA=-g(sin theta-mu cos theta)(n-1)` `g sin theta[1+n-1]=mu g cos theta(n-1)` `mu = TAN theta[(n)/(n-1)]`. |
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