A force of 400 N acting horizontal pushes u pa 20 kg block placed on a rough inclined plane which makes an angle of `45^(@)` with the horizontal. The acceleration experienced by the block is `0.6 m//s^(2)`.Find the coefficient of sliding friction between the box and incline.

Explanation

The horizontally directed force 400N and weight 20kg of the block are resolved into two mutually perpendicular components, parallel and perpendicular to the plane as shown.
`N = 20 g cos 45^(@) + 400 sin 45^(@) =421.4N`
The frictional force experienced by the block
`F = mu N = mu xx 421.4 = 421.4 muN`
As the accelerated motion is taking placed up the plane.
`400 cos 45^(@) - 20 g sin 45^(@) -f =20 a `

`( 400)/(sqrt(2)) - (20 xx 9.8)/( sqrt(2)) -421.4mu =20a = 20 xx 0.6 =12 `
`mu = ((400)/( sqrt(2))=- ( 196)/(sqrt(2))-12)xx(1)/(421.4)`
`= (282.8-138.6 -12)/( 421.4)= 0.3137`
The coefficient of sliding friction between the block and the incline `=0.3137`