A body slides down an inclined plane of inclination $θ$ . The coefficient of friction down the plane varies in direct proportion to the distance moved down the plane (µ = k x). The body will move down the plane with a

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Constant acceleration = g sin $θ$

Constant acceleration = (g sin $θ$ - µ g cos $θ$ )

Constant retardation = (µ g cos $θ$ - g sin $θ$ )

Variable acceleration that first decreases from g sin $θ$ to zero and then becomes negative.

answer is D.

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Detailed Solution

Force of friction is

f = µR =µ mgcos $θ$ .

When the body slides down, the downward force along the plane = component mg sin $θ$ of the weight mg. Since the force of friction acts upwards along the plane, the effective downward force = mg sin $θ$ - µmg cos $θ$ = mg (sin $θ$ - µ cos $θ$ )

$∴$ Acceleration = force / mass = $g (\sin θ - µ \cos θ) = g (\sin θ - k x \cos θ)$ . Hence, the acceleration varies with x and decreases as x increases.