A body of mass m is pushed with the initial velocity v_(0) up an inclined plane set at an angle theta to the horizontal. The fricitional coefficient is equal to mu. What distance will the body cover before it stops and what work do the frictional forces perform over this distance ?

A body of mass m is pushed with the initial velocity v_(0) up an incli

1.	A body of mass m is pushed with the initial velocity v_(0) up an inclined plane set at an angle theta to the horizontal. The fricitional coefficient is equal to mu. What distance will the body cover before it stops and what work do the frictional forces perform over this distance ?
Answer» SOLUTION :The net force on the body in the downward direction is given by `m g SIN theta + mu m g cos theta` `THEREFORE` retardating `a = - g(sin theta + mu cos theta)` Applying the formula `v^(2) = u^(2) + 2as`, we have , `0 = v_(0)^(2) - 2g(sin theta + mu cos theta)s or s = (v_(0)^(2))/(2g (sin theta + mu cos theta))`. WORK done by frictional force `= - mu m g cos theta xx s` `therefore W_(f_(r)) =-(mu m g cos theta xx v_(0)^(2))/(2g (sin theta + mu cos theta))=-(mumg v_(0)^(2)cos theta)/(2(sin theta + mu cos theta)g) = - (mu m v_(0)^(2))/(2(TAN theta + mu))`.