A body of mass m is released from the top of a rough inclined plane If the frictional force - Physics - Laws Of Motion

Question

A body of mass m is released from the top of a rough inclined
plane If the frictional force be fk Then prove that the
body will reach the bottom with velocity given by v= under root
2/m(mgh - fkL)

Panchali Dutta · Accepted Answer

Assuming h is the height of the inclined plane and L is the length of the inclined plane

For body to slide, F = mgsinÎ¸ - fk â‡’a = gsinÎ¸-fk/m We know, $v^{2} = u^{2} + 2 a L v^{2} = 0^{2} + [g \sin θ - \frac{f_{k}}{m}] 2 L = \frac{2}{m} [m g \sin θ - f_{k}] L \sin θ = \frac{h}{L} \Rightarrow v^{2} = \frac{2}{m} [m g h - f_{k} L]$

A body of mass m is released from the top of a rough inclined plane. If the frictional force be fk. Then prove that the body will reach the bottom with velocity given by v= under root 2/m(mgh - fkL)

A body of mass m is released from the top of a rough inclined plane. If the frictional force be f_k.Then prove that the body will reach the bottom with velocity given by v= under root 2/m(mgh - f_kL)