A wooden box of mass 8 kg slides down an inclined plane of inclination 30${^\circ}$ to the horizontal with a constant acceleration of $0.4\ ms^{- 2}$. What is the force of friction between the box and inclined plane? [$\ g = 10\ ms^{- 2}$]

Question

Dr. Abubacker · Accepted Answer

(a) $ma = mg\sin{	heta - f}$
or $f = mg\sin{	heta - ma}$
$= 8\left\lbrack 10 	imes \frac{1}{2} - 0.4 ightbrack$ N = 8$	imes 4.6$N = 36.8 N

Q. A wooden box of mass 8 kg slides down an inclined plane of inclination 30${^\circ}$ to the horizontal with a constant acceleration of $0.4\ ms^{- 2}$. What is the force of friction between the box and inclined plane? [$\ g = 10\ ms^{- 2}$]

Explanation

Reviewed by Dr. Abubacker

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