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

Let h be the height of inclined plane `Rightarrow h = 5 sin 37^(@) = 3 m`
(a) As the block slides down the inclined plane, it losses GPE and gains KE. As there is only a conservative force present, so loss of GPE = gain of KE = mgh = `1/2 mv^(2)`
`Rightarrow v = sqrt(2gh) = sqrt(2 xx9.8 xx 3) = 7.67 m/s.`
(b) As the block comes down, it loses GPE. It gains KE and also does work aggainst friction. Loss in `GPE = gain in KE + work done against friction`
`Rightarrow mgh = (1/2 mv^(2) -0) + (mu mg cos 37^(@))s`
`Rightarrow 3mg = 1/2 mv^(2) + (0.25) xx mg xx 4/5 xx5`
`Rightarrow v = sqrt 4g = 6.26 m/s`