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(B) The forces acting on the block along the inclined plane are the component of gravity $mg \sin 	heta$ acting downwards and the frictional force $f

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$\sqrt{2gl(\sin 	heta - \mu \cos 	heta)}$

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(B) The forces acting on the block along the inclined plane are the component of gravity $mg \sin 	heta$ acting downwards and the frictional force $f_k = \mu N = \mu mg \cos 	heta$ acting upwards.Applying Newton's second law along the incline: $mg \sin 	heta - \mu mg \cos 	heta = ma$.Thus,the acceleration of the block is $a = g(\sin 	heta - \mu \cos 	heta)$.Using the kinematic equation $v^2 = u^2 + 2as$,where initial velocity $u = 0$ and distance $s = l$:$v^2 = 0 + 2 \cdot g(\sin 	heta - \mu \cos 	heta) \cdot l$.Therefore,the velocity at the bottom is $v = \sqrt{2gl(\sin 	heta - \mu \cos 	heta)}$.

$A$ block is kept on an inclined plane of inclination $\theta$ and length $l$. The velocity of the block at the bottom of the inclined plane is (the coefficient of friction is $\mu$):

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