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  4. A wooden box of mass 8 kg slides down an inclined plane of inclination 30° 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]

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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? $\left[g=10 \,ms ^{-2}\right]$

Laws of Motion

Solution:

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$m g \sin \theta-\mu N=m a$
$\Rightarrow \quad \mu N=m g \sin \theta-m a$
$=(8 kg )\left(10 \frac{ m }{ s ^{2}}\right) \sin 30^{\circ}-(8 kg )\left(0.4 ms ^{-2}\right)=40 N -3.2 N$
$\Rightarrow $ Frictional force $=36.8 N$