A block of mass 'm' with an initial kinetic e | vedclass

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(D) Let the body slide up to a distance '$s$' before coming to rest.The height reached is $h = s \sin 	heta$.According to the work-energy theorem, th

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

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(D) Let the body slide up to a distance '$s$' before coming to rest.The height reached is $h = s \sin 	heta$.According to the work-energy theorem, the total work done by all forces equals the change in kinetic energy:$W_{friction} + W_{gravity} = K_f - K_i$$-\mu mg \cos 	heta \cdot s - mg \sin 	heta \cdot s = 0 - E$$\mu mgs \cos 	heta + mgs \sin 	heta = E$$s(mg(\mu \cos 	heta + \sin 	heta)) = E$$s = \frac{E}{mg(\mu \cos 	heta + \sin 	heta)}$The work done against friction is $W_{against\ friction} = |W_{friction}| = \mu mg \cos 	heta \cdot s$.Substituting the value of '$s$':$W_{against\ friction} = \mu mg \cos 	heta \cdot \left( \frac{E}{mg(\mu \cos 	heta + \sin 	heta)} ight)$$W_{against\ friction} = \frac{\mu E \cos 	heta}{\sin 	heta + \mu \cos 	heta}$

$A$ block of mass '$m$' with an initial kinetic energy '$E$' moves up an inclined plane of inclination '$\theta$'. If '$\mu$' is the coefficient of friction between the plane and the body, the work done against friction before coming to rest is

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