• # question_answer A body slides down an inclined plane of inclination 6. The coefficient of friction down the plane varies in direct proportion to the distance moved down the plane $(\mu =k\,x)$. The body will move down the plane with a A) constant acceleration$~=g\text{ }sin\text{ }\theta$ B) constant acceleration$=(g\,\sin \,\theta -\mu \,g\,\cos \theta )$ C) constant retardation $=(\mu \,g\,\,\cos \,\theta -g\sin \theta )$ D) variable acceleration that first decreases from $g\text{ }sin\text{ }\theta$ to zero and then becomes negative.

 Force of friction is $F=\mu R=\mu \,mg\cos \theta$ When the body slides down, the downward force along the plane = component $mg\sin \theta$of the weight mg. since the force of friction acts upwards along the plane, the effective downward force = $mg\sin \theta -\mu mg\cos \theta =mg(sin\theta -\mu cos\theta )$ $\therefore$Acceleration = force/ mass = $g(sin\theta -\mu cos\theta )$ $=g(\sin \theta -kx\cos \theta )$ Hence, the acceleration varies with x and decreases as x increases. Thus, the correct choice is [D].