Work, Energy and Power: Exam-Pattern Drill (8)

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Tags Work Energy Power

Question

A 22 kg block slides down a rough inclined plane of length 55 m and inclination 30°30°. The coefficient of kinetic friction is μk=0.2\mu_k = 0.2. Using the work-energy theorem, find the speed of the block at the bottom. Take g=10g = 10 m/s2^2.

Solution — Step by Step

Three forces act on the block: gravity, normal reaction, and kinetic friction. Normal reaction is perpendicular to motion — it does zero work. Only gravity and friction matter for the work-energy theorem.

The block descends a vertical height h=Lsin⁡30°=5×0.5=2.5h = L \sin 30° = 5 \times 0.5 = 2.5 m.

Wgravity=mgh=2×10×2.5=50 JW_{\text{gravity}} = mgh = 2 \times 10 \times 2.5 = 50 \text{ J}

Normal force on incline: N=mgcos⁡30°=2×10×32=103N = mg \cos 30° = 2 \times 10 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} N.

Friction force f=μkN=0.2×103=23f = \mu_k N = 0.2 \times 10\sqrt{3} = 2\sqrt{3} N. Friction opposes motion, so

Wfriction=−fL=−23×5=−103≈−17.32 JW_{\text{friction}} = -fL = -2\sqrt{3} \times 5 = -10\sqrt{3} \approx -17.32 \text{ J} ΔKE=Wnet=50−103\Delta KE = W_{\text{net}} = 50 - 10\sqrt{3} 12(2)v2=50−17.32=32.68  ⟹  v2=32.68\frac{1}{2}(2)v^2 = 50 - 17.32 = 32.68 \implies v^2 = 32.68 v≈5.72 m/sv \approx 5.72 \text{ m/s}

Final Answer: v≈5.72v \approx 5.72 m/s.

Why This Works

The work-energy theorem says ΔKE=Wnet\Delta KE = W_{\text{net}}, regardless of path. It bypasses Newton’s laws and time-of-motion calculations entirely. For incline problems with friction, this is usually the fastest route — no need to find acceleration first and then use kinematics.

The key insight: gravity does positive work because force and displacement both have a downward component along the incline. Friction always does negative work on the block because it opposes motion.

Alternative Method

Energy conservation with the friction loss term: mgh=12mv2+μkmgcos⁡θ⋅Lmgh = \frac{1}{2}mv^2 + \mu_k mg\cos\theta \cdot L. Same equation, same answer. Use whichever framing feels natural — they’re algebraically identical.

Students often forget the cos⁡θ\cos\theta in the normal force on an incline. They use N=mgN = mg (the flat-ground formula) and get a friction force that’s too large. Always draw the FBD and resolve mgmg into components along and perpendicular to the incline.

JEE Main 2023 Shift 2 had nearly this exact setup with different numbers. NEET tends to ask the same question but stops at “find acceleration” instead of “find final speed”. Same physics, one less step.

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