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September 2021 2 35 Report

A block is released from the top of an inclined plane of inclination theta. Its velocity at the bottom of the plane is v. If it slides down a rough inclined plane of same inclination its velocity on reaching the bottom is v/2. The coefficient of friction is.
A)3/4 cot theta
B) 1/4 cot theta
C)1/4 tan theta
D)3/4 tan theta

phillipinestest

Answer: D) 3 / 4 $tan \theta$

From the data given, we have initial velocity as u = 0 and final velocity v = v and acceleration a = g sin ? and displacement s = x thereby equation of motion,

$v ^2 = u ^2+2 a s \\=> v 2 = 2 g sin\theta.$

And for the sliding down motion, initial velocity u = 0 and final velocity v = v / 2, acceleration,

$a = ( sin \theta - \mu cos \theta ) g. v^2 \\= u^2+2 a s \\=> v^2 / 4 = 2 x g ( sin \theta - \mu cos \theta ) x. \\Thereby, \mu = sin \theta / ( sin \theta - \mu cos \theta ) \\=> \mu sin \theta - \mu cos \theta = sin \theta \\=> 3 sin \theta = 4 \mu cos \theta \\=> \mu = 3 / 4 tan \theta$

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