A wooden box of mass `8kg` slides down an inclined plane of inclination `30^(@)` to the horizontal with a constant acceleration of `0.4ms^(-2)` What is the force of friction between the box and inclined plane ? `(g = 10m//s^(2))` .

Options

A. `36.8N`
B. `65.6N`
C. `76.8N`
D. `63.8N`

Answer

a

Explanation

Here, `m = 8kg, theta = 30^(@), a = 0.4m//s^(2), F = ?`
`a =g sin theta -(F)/(m)`
`0.4 =10sin30^(@) - (F)/(8)`
`(F)/(8) =10xx(1)/(2)- 0.4 =4.6`
`F = 4.6 xx 8 N=36.8N` .