Work, Energy and Power: Common Mistakes and Fixes (3)

hard • 3 min read
Tags Work Energy Power

Question

A 2 kg2 \text{ kg} block slides down a rough incline of angle 30°30° and length 5 m5 \text{ m}. The coefficient of kinetic friction is μk=0.2\mu_k = 0.2. Three students try to find the speed at the bottom and each makes a different mistake. Identify the mistakes and find the correct speed. Take g=10 m/s2g = 10 \text{ m/s}^2.

Solution — Step by Step

The block starts from rest, so by the work-energy theorem:

12mv2=Wgravity+Wfriction\tfrac{1}{2}mv^2 = W_{\text{gravity}} + W_{\text{friction}}

Height descended: h=Lsin⁡θ=5×0.5=2.5 mh = L\sin\theta = 5 \times 0.5 = 2.5 \text{ m}. So Wgravity=mgh=2×10×2.5=50 JW_{\text{gravity}} = mgh = 2 \times 10 \times 2.5 = 50 \text{ J}.

Friction force: f=μkN=μkmgcos⁡θ=0.2×2×10×cos⁡30°=4×0.866≈3.46 Nf = \mu_k N = \mu_k mg\cos\theta = 0.2 \times 2 \times 10 \times \cos 30° = 4 \times 0.866 \approx 3.46 \text{ N}. Work done by friction: Wf=−fL=−3.46×5=−17.3 JW_f = -fL = -3.46 \times 5 = -17.3 \text{ J}.

12(2)v2=50−17.3=32.7 J\tfrac{1}{2}(2)v^2 = 50 - 17.3 = 32.7 \text{ J}

v2=32.7  ⟹  v≈5.72 m/sv^2 = 32.7 \implies v \approx 5.72 \text{ m/s}

Mistake A: Student writes Wf=−μkmgLW_f = -\mu_k mg L (forgetting the cos⁡θ\cos\theta in normal force on an incline). They get Wf=−20 JW_f = -20 \text{ J} and v≈5.48 m/sv \approx 5.48 \text{ m/s} — close but wrong.

Mistake B: Student takes h=L=5 mh = L = 5 \text{ m} (treating the incline length as height). They get Wg=100 JW_g = 100 \text{ J} and a wildly inflated vv.

Mistake C: Student adds friction work as positive (sign error). They get 12mv2=67.3 J\tfrac{1}{2}mv^2 = 67.3 \text{ J} and v≈8.2 m/sv \approx 8.2 \text{ m/s} — friction never speeds anything up.

Why This Works

The work-energy theorem is the cleanest tool for incline problems because it skips the time variable entirely. Two cautions: always project gravity onto the height drop (not the incline length), and friction always does negative work when the block slides down (force opposes motion).

On an incline, N=mgcos⁡θN = mg\cos\theta, not mgmg. This is the single most-tested point in JEE Main friction problems — the cosine factor sneaks into every friction force on tilted surfaces.

Alternative Method

Use Newton’s second law along the incline. Net force down the slope: mgsin⁡θ−μkmgcos⁡θ=10−3.46=6.54 Nmg\sin\theta - \mu_k mg\cos\theta = 10 - 3.46 = 6.54 \text{ N}. Acceleration: a=3.27 m/s2a = 3.27 \text{ m/s}^2. Then v2=2aL=2×3.27×5=32.7v^2 = 2aL = 2 \times 3.27 \times 5 = 32.7, giving v≈5.72 m/sv \approx 5.72 \text{ m/s}. Same answer — energy method is just faster when no time data is needed.

Common Mistake

The most damaging error in WEP problems is forgetting that only the component of force along displacement does work. On inclines, that means cos⁡θ\cos\theta for friction (because friction is along the incline) but the height drop Lsin⁡θL\sin\theta for gravity (because gravity is vertical). Students mix these up under exam pressure. Sketch the FBD, mark the height hh separately from the path length LL, and the trigonometry sorts itself out.

Final answer: v≈5.72 m/sv \approx 5.72 \text{ m/s} at the bottom.

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