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Question

(A) When a body moves up an inclined plane,the forces acting against the motion are the component of gravity along the plane $(mg \sin 	heta)$ and th

Vedclass · Accepted Answer

$\left(\frac{2+\sqrt{3}}{4}\right) g$

Vedclass · Answer

(A) When a body moves up an inclined plane,the forces acting against the motion are the component of gravity along the plane $(mg \sin 	heta)$ and the frictional force $(f = \mu N = \mu mg \cos 	heta)$.According to Newton's second law,the net force $F_{net} = ma = -(mg \sin 	heta + \mu mg \cos 	heta)$.Thus,the retardation $a$ is given by $a = g(\sin 	heta + \mu \cos 	heta)$.Substituting the given values $	heta = 30^{\circ}$ and $\mu = 0.5$:$a = g(\sin 30^{\circ} + 0.5 \cos 30^{\circ})$$a = g\left(\frac{1}{2} + 0.5 	imes \frac{\sqrt{3}}{2}ight)$$a = g\left(\frac{1}{2} + \frac{\sqrt{3}}{4}ight)$$a = g\left(\frac{2 + \sqrt{3}}{4}ight)$.

$A$ body is made to move up along an inclined plane of inclination $30^{\circ}$ and the coefficient of friction is $0.5$. Then its retardation is ($g$ = acceleration due to gravity).

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