Work, Energy and Power: PYQ Walkthrough (4)

easy • 3 min read
Tags Work Energy Power

Question

A block of mass 22 kg is pushed up a smooth incline of angle 30°30° by a horizontal force of 2020 N over a displacement of 55 m along the incline. Calculate the work done by (a) the applied force, (b) gravity, and (c) the net work done on the block. (Take g=10g = 10 m/s².) This pattern appeared in NEET 2022 and JEE Main 2023.

Solution — Step by Step

The applied force is horizontal but the displacement is along the incline. The component of the 2020 N force along the incline (up the slope) is Fcos⁡θ=20cos⁡30°=20×32=103F\cos\theta = 20 \cos 30° = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} N.

Wapplied=Fcos⁡θ⋅s=103×5=503≈86.6 JW_{\text{applied}} = F\cos\theta \cdot s = 10\sqrt{3} \times 5 = 50\sqrt{3} \approx 86.6 \text{ J}

Gravity acts vertically downward. The vertical displacement of the block is h=ssin⁡θ=5×0.5=2.5h = s\sin\theta = 5 \times 0.5 = 2.5 m (upward).

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

The negative sign tells us gravity opposes the motion.

The normal reaction does zero work (perpendicular to displacement). The incline is smooth, so no friction.

Wnet=Wapplied+Wgravity=503−50≈36.6 JW_{\text{net}} = W_{\text{applied}} + W_{\text{gravity}} = 50\sqrt{3} - 50 \approx 36.6 \text{ J}

Final answers: Wapplied=503W_{\text{applied}} = 50\sqrt{3} J, Wgravity=−50W_{\text{gravity}} = -50 J, Wnet≈36.6W_{\text{net}} \approx 36.6 J.

Why This Works

Work depends on the angle between force and displacement, not on the absolute direction of either. A horizontal force on an inclined surface always has a non-trivial component along the incline — students who set W=F⋅sW = F \cdot s without resolving lose the question instantly.

The work-energy theorem says Wnet=ΔKEW_{\text{net}} = \Delta KE, so this 36.6 J is exactly the kinetic energy gained by the block. We could verify with v2=u2+2asv^2 = u^2 + 2as if needed.

Alternative Method

Use energy conservation directly. The block gains potential energy mgh=50mgh = 50 J and kinetic energy ΔKE\Delta KE. Energy input by applied force = 50350\sqrt{3} J. So ΔKE=503−50≈36.6\Delta KE = 50\sqrt{3} - 50 \approx 36.6 J. Same answer, fewer steps.

Students often write W=F⋅s=20×5=100W = F \cdot s = 20 \times 5 = 100 J and stop. This ignores the cos⁡θ\cos\theta factor between force direction and displacement direction. Always resolve forces along the displacement.

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