A block slides down a rough incline — find acceleration using Newton's laws

Question

A block of mass $5\,\text{kg}$ slides down a rough inclined plane of angle $30°$ with the horizontal. The coefficient of kinetic friction between the block and the surface is $\mu_k = 0.2$ . Find the acceleration of the block. (Take $g = 10\,\text{m/s}^2$ )

(JEE Main 2022, similar pattern)

Solution — Step by Step

Forces on the block along the incline:

Down the incline: Component of gravity $= mg\sin\theta = 5 \times 10 \times \sin 30° = 25\,\text{N}$
Up the incline: Friction $= \mu_k N$

Normal force: $N = mg\cos\theta = 5 \times 10 \times \cos 30° = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3}\,\text{N}$

f = \mu_k N = 0.2 \times 25\sqrt{3} = 5\sqrt{3} \approx 8.66\,\text{N}

Net force down the incline:

F_{net} = mg\sin\theta - \mu_k mg\cos\theta = mg(\sin\theta - \mu_k\cos\theta)

a = g(\sin\theta - \mu_k\cos\theta)

a = 10\left(\sin 30° - 0.2\cos 30°\right) = 10\left(0.5 - 0.2 \times 0.866\right)

a = 10(0.5 - 0.1732) = 10 \times 0.3268

\boxed{a \approx 3.27\,\text{m/s}^2}

Why This Works

On an incline, gravity has two components: $mg\sin\theta$ pulls the block down the slope, and $mg\cos\theta$ pushes it into the surface (creating the normal force). Friction opposes motion, so it acts up the slope when the block slides down.

The general formula $a = g(\sin\theta - \mu_k\cos\theta)$ is worth remembering. It tells us:

If $\tan\theta > \mu_k$ : the block accelerates down (our case: $\tan 30° = 0.577 > 0.2$ )
If $\tan\theta = \mu_k$ : the block moves at constant velocity
If $\tan\theta < \mu_k$ : the block doesn’t slide at all (static friction is enough)

Alternative Method — Energy approach

Over a distance $s$ down the incline:

Work done by gravity: $mgs\sin\theta$ Work done by friction: $-\mu_k mgs\cos\theta$

By work-energy theorem: $\frac{1}{2}mv^2 = mgs(\sin\theta - \mu_k\cos\theta)$

Using $v^2 = 2as$ : $a = g(\sin\theta - \mu_k\cos\theta)$ . Same result.

Always check whether the block actually slides before computing acceleration. The condition for sliding is $mg\sin\theta > \mu_s mg\cos\theta$ , i.e., $\tan\theta > \mu_s$ (use static friction coefficient here). If the block doesn’t slide, $a = 0$ regardless of the angle.

Common Mistake

A very common error: using $mg$ as the normal force instead of $mg\cos\theta$ . On an incline, the normal force is $N = mg\cos\theta$ , NOT $mg$ . The full weight $mg$ acts vertically, but the surface pushes back only against the perpendicular component. Using $N = mg$ overestimates friction and gives the wrong acceleration.

Question

Solution — Step by Step

Why This Works

Alternative Method — Energy approach

Common Mistake

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