A 20kg box is gently placed on a rough inclined plane of inclination `30^(@)` with horizontal. The coefficient of sliding friction between the box and the plane is 0.5.Find the acceleration of the box down the incline.

Explanation

In solving inclined plane problems, theX and Y directions along which the forces are to be considered, may be taken as shown. The components of weight of the box are
(i) mg `sin alpha` acting down the plane and
(ii) `mg cos alpha` acting perpendicular to the plane.
`N =mg cos alpha`
`m g sin alpha - mu N = ma `
`rArr mg sin alpha - mu mg cos alpha = ma `
`a =g sin alpha - mu g cosalpha - g ( sin alpha - mu cos alpha ) `
` =9.8 ((1)/(2) - 0.5 xx( sqrt(3))/( 2)) = 4.9xx 0.3072 =1.505 m//s^(2)`
The box accelerates down the plane at `1.505m//s^(2)`.