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(D) Let $l$ be the total length of the inclined plane and $	heta$ be the angle of inclination.According to the work-energy theorem,the total work don

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$	an^{-1}\left[\left(\frac{n-1}{n}ight) \mu_kight]$

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(D) Let $l$ be the total length of the inclined plane and $	heta$ be the angle of inclination.According to the work-energy theorem,the total work done on the body is equal to the change in its kinetic energy.Since the body starts from rest and comes to rest at the bottom,the change in kinetic energy $\Delta K = 0$.The forces acting on the body are gravity,friction,and the normal force.$1$. Work done by gravity $(W_g)$: $W_g = mgh = mg(l \sin 	heta)$.$2$. Work done by friction $(W_f)$: Friction acts only on the lower part of length $l(1 - 1/n)$. Thus,$W_f = -f_k \cdot d = -(\mu_k mg \cos 	heta) \cdot l(1 - 1/n)$.$3$. Work done by the normal force $(W_N)$: Since the normal force is always perpendicular to the displacement,$W_N = 0$.Applying the work-energy theorem: $W_g + W_f + W_N = 0$.$mg l \sin 	heta - \mu_k mg \cos 	heta \cdot l \left(\frac{n-1}{n}ight) = 0$.Dividing by $mg l \cos 	heta$ (assuming $\cos 	heta 
eq 0$):$	an 	heta = \mu_k \left(\frac{n-1}{n}ight)$.Therefore,$	heta = 	an^{-1}\left[\left(\frac{n-1}{n}ight) \mu_kight]$.

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