Question

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))` .
A. 36.8N
B. 76.8 N
C. 65.6 N
D. none of these

Answer 1

Correct Answer - A
(a)`mg sin theta-f=ma`(f=force of friction)
or `" "f=m(g sin theta-a)`
`=8(10xx(1)/(2)-0.4)=36.8N`