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

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Question

(D) The total retarding force acting on the body moving up the inclined plane is $F_{net} = mg \sin 	heta + f_k$, where $f_k = \mu R = \mu mg \cos

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The total retarding force acting on the body moving up the inclined plane is $F_{net} = mg \sin 	heta + f_k$, where $f_k = \mu R = \mu mg \cos 	heta$.Thus, $F_{net} = mg(\sin 	heta + \mu \cos 	heta)$.The retardation $a$ is given by $a = \frac{F_{net}}{m} = g(\sin 	heta + \mu \cos 	heta)$.Using the work-energy theorem, the total work done by all forces equals the change in kinetic energy: $W_{total} = \Delta KE = 0 - E = -E$.The work done by gravity is $W_g = -mg \sin 	heta \cdot s$ and the work done against friction is $W_f = f_k \cdot s = \mu mg \cos 	heta \cdot s$.From $v^2 - u^2 = 2as$, we have $0 - u^2 = -2as$, so $s = \frac{u^2}{2a} = \frac{2E/m}{2g(\sin 	heta + \mu \cos 	heta)} = \frac{E}{mg(\sin 	heta + \mu \cos 	heta)}$.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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