A block is placed on the top of a plane inclined at `37^@` with horizontal. The length of the plane is `5m`. The block slides down the plane and reaches the bottom.

a. Find the speed of the block at the bottom if the inclined plane is smooth.
b. Find the speed of the block at the bottom if the coefficient of friction is `0.25`.

Explanation

When the block slides down on the smooth wedge, the wedge moves backwards. In the horizontal direction there is no external force. `vecF_(x)=0` , `therefore vecP_(x)` = constant
`vecP_(f) = vecP_(i)` (along x-axis) `mvecu + MvecV =0`
`x_(1)` = forward distance moved by the block along x-axis.
`x_2` = backward distance moved by the wedge along x-axis.
`mvecu=-MvecV, m(x_(1)/t) =Mx_(2)/t`
`mx_(1) =Mx_(2).x_(1) =(ML)/(M+m) =(Ml cos theta)/(M+m)`
`x_(2) =(mL)/(M+m) =(mlcos theta)/(M+m)`
L can also be written as `l cos theta`