JEE Main Physics: Acceleration on Incline Concept 💡

 

❓ Question

A cubic block of mass $m$ is sliding down an inclined plane at $60^\circ$ with an acceleration of $g/2$.

Find the coefficient of kinetic friction $\mu_k$.

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🖼️ Question Image

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✍️ Short Solution

This is a direct Newton’s Second Law + friction problem.
No tricks. Just resolve forces properly 🔥

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🔹 Step 1 — Draw Free Body Diagram (MOST IMPORTANT 💯)**

For block on incline at angle $60^\circ$:

Forces acting:

• Weight $mg$ (vertical downward)

• Normal reaction $N$
  

• Kinetic friction $f_k = \mu_k N$ (up the plane)

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🔹 Step 2 — Resolve Weight into Components

Along incline:

$$mg \sin 60^\circ$$

Perpendicular to incline:

$$mg \cos 60^\circ$$

So normal reaction:

$$N = mg \cos 60^\circ$$

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🔹 Step 3 — Apply Newton’s Second Law (Along Incline)

Net force down the plane:

$$mg \sin 60^\circ - f_k = ma$$

Since:

$$f_k = \mu_k N = \mu_k mg \cos 60^\circ$$

So:

$$mg \sin 60^\circ - \mu_k mg \cos 60^\circ = m\frac{g}{2}$$

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🔹 Step 4 — Cancel Common Terms

Divide entire equation by $mg$:

$$\sin 60^\circ - \mu_k \cos 60^\circ = \frac{1}{2}$$

Substitute values:

$$\sin 60^\circ = \frac{\sqrt{3}}{2}$$ $$\cos 60^\circ = \frac{1}{2}$$

So:

$$\frac{\sqrt{3}}{2} - \mu_k \frac{1}{2} = \frac{1}{2}$$

Multiply by 2:

$$\sqrt{3} - \mu_k = 1$$

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🔹 Step 5 — Solve for $\mu_k$

$$\mu_k = \sqrt{3} - 1$$

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✅ Final Answer

$$\boxed{\mu_k = \sqrt{3} - 1}$$

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⭐ Golden JEE Insight

On an incline:

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

🧠 If acceleration given, directly use:

$$\mu_k = \frac{\sin\theta - \frac{a}{g}}{\cos\theta}$$

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🔥 Quick Shortcut Formula

For sliding block:

$$\mu_k = \frac{\sin\theta - \frac{a}{g}}{\cos\theta}$$

Put:

$$\theta=60^\circ,\quad a=\frac{g}{2}$$

Instant answer.