A body is moving up an inclined plane of angl | vedclass

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(D) The forces acting on the body moving up the inclined plane are the component of gravity $mg \sin 	heta$ acting down the plane and the frictional

Vedclass · Accepted Answer

$\frac{\mu E \cos 	heta}{\mu \cos 	heta+\sin 	heta}$

Vedclass · Answer

(D) The forces acting on the body moving up the inclined plane are the component of gravity $mg \sin 	heta$ acting down the plane and the frictional force $f = \mu R = \mu mg \cos 	heta$ acting down the plane.The total retarding force is $F_{net} = mg \sin 	heta + \mu mg \cos 	heta = mg(\sin 	heta + \mu \cos 	heta)$.The retardation (deceleration) is $a = \frac{F_{net}}{m} = g(\sin 	heta + \mu \cos 	heta)$.Using the work-energy theorem,the work done by all forces equals the change in kinetic energy:$W_{total} = \Delta K = K_f - K_i = 0 - E = -E$.The total work done is the sum of work done against gravity $(W_g)$ and work done against friction $(W_f)$:$W_g + W_f = -E$.Since $W_g = -mg \sin 	heta \cdot s$ and $W_f = -\mu mg \cos 	heta \cdot s$,where $s$ is the distance traveled along the plane:$s = \frac{v^2}{2a} = \frac{u^2}{2g(\sin 	heta + \mu \cos 	heta)} = \frac{2E/m}{2g(\sin 	heta + \mu \cos 	heta)} = \frac{E}{mg(\sin 	heta + \mu \cos 	heta)}$.The work done against friction is $W_f = \mu mg \cos 	heta \cdot s = \mu mg \cos 	heta \cdot \frac{E}{mg(\sin 	heta + \mu \cos 	heta)} = \frac{\mu E \cos 	heta}{\sin 	heta + \mu \cos 	heta}$.

$A$ body is moving up an inclined plane of angle $\theta$ with an initial kinetic energy $E$. The coefficient of friction between the plane and the body is $\mu$. The work done against friction before the body comes to rest is

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