Work, Energy and Power: JEE Main 2026 PYQs

Answer: 35

Key idea: Work done by net force using dot product of force and displacement vectors

1. Find net force: $\vec{F_{net}} = (2\hat{i}+3\hat{j}+4\hat{k}) + (3\hat{i}-\hat{j}-2\hat{k}) = 5\hat{i} + 2\hat{j} + 2\hat{k}$ N
2. Find unit vector along displacement: $\hat{d} = \frac{3\hat{i}-4\hat{j}}{\sqrt{3^2+4^2}} = \frac{3\hat{i}-4\hat{j}}{5}$
3. Calculate displacement vector: $\vec{s} = 25\hat{d} = 25 \times \frac{3\hat{i}-4\hat{j}}{5} = 15\hat{i} - 20\hat{j}$ m
4. Apply work formula: $W = \vec{F_{net}} \cdot \vec{s} = (5\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (15\hat{i} - 20\hat{j})$
5. Calculate dot product: $W = 5(15) + 2(-20) + 2(0) = 75 - 40 = 35$ J

Answer: 200

Key idea: Power as dot product of force and velocity; Integration to find velocity from time-dependent force

1. Find acceleration: $\vec{a} = \frac{\vec{F}}{m} = \frac{2t\hat{i} + 6t^2\hat{j}}{2} = (t\hat{i} + 3t^2\hat{j})$ m/s²
2. Integrate to find velocity: $\vec{v} = \int \vec{a} \, dt = (\frac{t^2}{2}\hat{i} + t^3\hat{j})$ m/s (using initial condition $\vec{v}(0) = 0$)
3. At $t = 2$ s: $\vec{v} = (2\hat{i} + 8\hat{j})$ m/s
4. At $t = 2$ s: $\vec{F} = (4\hat{i} + 24\hat{j})$ N
5. Calculate power: $P = \vec{F} \cdot \vec{v} = 4(2) + 24(8) = 8 + 192 = 200$ W

Answer: Option (1)

Key idea: Work done by friction on an object moving with an inclined plane assembly

1. Given: mass $m = 1$ kg on inclined plane ($\theta = 30°$), assembly moves up with constant velocity $v = 4$ m/s for $t = 2$ s
2. Since mass is at rest on the plane and assembly moves with constant velocity, friction balances the weight component: $f = mg\sin(30°) = 1 \times 10 \times 0.5 = 5$ N (up the plane)
3. Vertical displacement of mass: $s = vt = 4 \times 2 = 8$ m
4. Component of displacement along the inclined plane: $s_{\parallel} = s\sin(30°) = 8 \times 0.5 = 4$ m
5. Work done by friction: $W = f \times s_{\parallel} = 5 \times 4 = 20$ J

Answer: Option (4)

Key idea: Work-Energy Theorem with Non-Conservative Forces

1. Convert units: $m = 1\text{ g} = 0.001\text{ kg}$, $h = 1\text{ km} = 1000\text{ m}$
2. Calculate work by gravity: $W_g = mgh = 0.001 \times 10 \times 1000 = 10\text{ J}$
3. Calculate initial KE: $KE_i = \frac{1}{2}mu^2 = 0$ (starts from rest)
4. Calculate final KE: $KE_f = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.001 \times 25 = 0.0125\text{ J}$
5. Apply work-energy theorem: $W_g + W_{resistive} = \Delta KE$
6. Solve: $10 + W_{resistive} = 0.0125$, so $W_{resistive} = -9.9875 \approx -9.98\text{ J}$

Answer: Option (3)

Key idea: Work-Energy Theorem

1. Given: mass $m = 1$ kg, velocity $v = 2x^2$, displacement from $x = 0$ to $x = 5$ m
2. Apply Work-Energy Theorem: $W = \Delta KE = KE_f - KE_i$
3. Find initial velocity at $x = 0$: $v_0 = 2(0)^2 = 0$ m/s
4. Find final velocity at $x = 5$: $v_5 = 2(5)^2 = 50$ m/s
5. Calculate $KE_i = \frac{1}{2}(1)(0)^2 = 0$ J and $KE_f = \frac{1}{2}(1)(50)^2 = 1250$ J
6. Work done $W = 1250 - 0 = 1250$ J

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