Question
The coefficient of friction between a body and the surface of an inclined plane at $45^{\circ}$ is $0.5$ if $g=9.8 \mathrm{~m} / \mathrm{s}^{2} .$ The acceleration of the body downwards in $\mathrm{m} / \mathrm{s}^{2}$ is(a) $\frac{4.9}{\sqrt{2}}$(b) $4.9 \sqrt{2}$(c) $19.2 \sqrt{2}$(d) $4.9$
Step 1
First, we need to find the gravitational force acting on the body along the inclined plane. Since the angle of inclination is $45^{\circ}$, the component of gravitational force along the inclined plane is $mg\sin{45^{\circ}}$. Show more…
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