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CGP EDU Academic Team
Published on: September 12, 2026
The coefficient of friction between a body and the surface of an inclined plane at $45^\circ$ is 0.5. If $g = 9.8 \, \mathrm{m/s^2}$ , the acceleration of the body downwards in $m/s^{2}$ is
Text Solution
Verified by ExpertsThe correct answer is:
A
a = $g(\sin \theta - \mu \cos \theta) = 9.8(\sin 45^\circ - 0.5 \cos 45^\circ)$
$= \frac{4.9}{\sqrt{2}} \mathrm{m/sec}^2$
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