Motion On An Inclined Plane

Question

A block slides down a rough incline tilted at 30 degrees to the horizontal where the coefficient of kinetic friction between block and surface is 0.2. Using g as 10 m/s^2, what is the acceleration of the block down the slope?

RankGuru · Accepted Answer

On a rough incline the component of gravity along the slope drives the block downward while kinetic friction opposes the motion up the slope. Resolving gravity, the driving component is $mg\sin	heta$ and the normal force is $mg\cos	heta$, so friction is $\mu mg\cos	heta$. Newton's Second Law along the incline gives $a = g(\sin	heta - \mu\cos	heta)$. Substituting, $a = 10(\sin 30^\circ - 0.2\cos 30^\circ) = 10(0.5 - 0.2 	imes 0.866) = 10(0.5 - 0.173) = 10(0.327) = 3.27	ext{ m/s}^2$. The 5.00 m/s^2 value ignores friction entirely and uses only $g\sin	heta$. The 1.73 m/s^2 value mistakenly uses $g\cos	heta$ scaled by friction alone. The 2.30 m/s^2 value comes from swapping sine and cosine in the two gravity components. A key conceptual point is that the acceleration here is independent of the block's mass, because both the driving gravity term and the friction term scale with mass, so it cancels throughout. Notice also that the block keeps sliding only because $	an	heta = 0.577$ exceeds $\mu = 0.2$; had friction been large enough that $\mu \ge 	an	heta$, the block would not accelerate at all. This is the standard NCERT inclined-plane analysis with friction. As a check, the answer is positive and smaller than the frictionless 5 m/s^2, confirming that friction reduces but does not reverse the downhill acceleration.

Motion On An Inclined Plane

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