Work, Energy and Power: Step-by-Step Worked Examples (6)

hard • 2 min read
Tags Work Energy Power

Question

A block of mass 2 kg2\ \text{kg} slides down a rough incline of angle 30∘30^\circ from rest. The coefficient of kinetic friction is μk=0.2\mu_k = 0.2. Using the work-energy theorem, find the speed after sliding 5 m5\ \text{m} along the slope. Take g=10 m/s2g = 10\ \text{m/s}^2.

Solution — Step by Step

Three forces act: gravity (component along slope), normal (perpendicular, no work), and kinetic friction (opposes motion). Only gravity and friction contribute to net work.

The component of weight along the slope is mgsin⁡θmg\sin\theta. Over s=5 ms = 5\ \text{m}:

Wg=mgsin⁡θ⋅s=2×10×0.5×5=50 JW_g = mg\sin\theta \cdot s = 2 \times 10 \times 0.5 \times 5 = 50\ \text{J}

Normal force N=mgcos⁡θ=2×10×32≈17.32 NN = mg\cos\theta = 2 \times 10 \times \tfrac{\sqrt{3}}{2} \approx 17.32\ \text{N}. Friction force f=μkN≈3.46 Nf = \mu_k N \approx 3.46\ \text{N}. Work done by friction (negative because it opposes motion):

Wf=−f⋅s=−3.46×5=−17.32 JW_f = -f \cdot s = -3.46 \times 5 = -17.32\ \text{J}

Wnet=ΔKEW_{\text{net}} = \Delta KE. Since the block starts from rest:

50−17.32=12(2)v2  ⟹  v2=32.68  ⟹  v≈5.72 m/s50 - 17.32 = \tfrac{1}{2}(2)v^2 \implies v^2 = 32.68 \implies v \approx 5.72\ \text{m/s}

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

Why This Works

The work-energy theorem replaces vector kinematics with a scalar energy balance. Whenever a problem mentions “rough surface”, “friction”, or “variable force”, reach for Wnet=ΔKEW_{\text{net}} = \Delta KE before trying F=maF = ma.

Energy methods also dodge the algebra of resolving acceleration vectors on inclines. We just track joules in and joules out.

Alternative Method

Find net acceleration using a=g(sin⁡θ−μkcos⁡θ)a = g(\sin\theta - \mu_k\cos\theta). Then v2=u2+2asv^2 = u^2 + 2as. Same answer, but you have to keep track of trigonometric signs throughout.

In JEE Advanced 2022, an inclined-plane PYQ had a spring at the bottom. Students who used energy conservation finished in 90 seconds; those who used Newton’s laws ran out of time.

Common Mistake

Students forget that WfW_f is negative. Plugging Wf=+17.32 JW_f = +17.32\ \text{J} gives v≈8.2 m/sv \approx 8.2\ \text{m/s}, a “trap” answer that is usually one of the MCQ options. Always write the sign explicitly when friction acts.

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