A body of 5, kg weight kept on a rough inclin | vedclass

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(A) When a body slides down an inclined plane with a constant velocity,the net force acting on the body is zero.This implies that the component of gra

Vedclass · Accepted Answer

$1/\sqrt{3}$

Vedclass · Answer

(A) When a body slides down an inclined plane with a constant velocity,the net force acting on the body is zero.This implies that the component of gravitational force acting down the plane is balanced by the force of kinetic friction.The component of weight acting down the plane is $mg \sin 	heta$.The force of kinetic friction is $f_k = \mu_k N = \mu_k mg \cos 	heta$.Equating these two forces: $mg \sin 	heta = \mu_k mg \cos 	heta$.Therefore,$\mu_k = \frac{\sin 	heta}{\cos 	heta} = 	an 	heta$.Given $	heta = 30^\circ$,we have $\mu_k = 	an 30^\circ = \frac{1}{\sqrt{3}}$.

$A$ body of $5\, kg$ weight kept on a rough inclined plane of angle $30^\circ$ starts sliding with a constant velocity. Then the coefficient of friction is:

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