Work, Energy & Power — Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Work, Energy and Power — work done by forces, the work-energy theorem, instantaneous power and energy conservation — with step-by-step solutions.
Solved JEE Main 2026 questions from the Work, Energy and Power chapter, covering work done by constant and time-dependent forces, the work-energy theorem with resistive forces, and instantaneous power, each with a concise step-by-step solution.
Solutions are AI-generated and pending review.
Solution
Net force:
$$\vec{F} = (2+3)\hat{i} + (3-1)\hat{j} + (4-2)\hat{k} = 5\hat{i} + 2\hat{j} + 2\hat{k}\ \text{N}$$Displacement of magnitude 25 m along $(3\hat{i} - 4\hat{j})$. Unit vector:
$$\hat{d} = \frac{3\hat{i} - 4\hat{j}}{\sqrt{3^2 + 4^2}} = \frac{3\hat{i} - 4\hat{j}}{5}$$$$\vec{s} = 25\,\hat{d} = 15\hat{i} - 20\hat{j}\ \text{m}$$Work done:
$$W = \vec{F}\cdot\vec{s} = (5)(15) + (2)(-20) + (2)(0) = 75 - 40 = 35\ \text{J}$$Answer: 35
Solution
Acceleration ($m = 2$ kg):
$$\vec{a} = \frac{\vec{F}}{m} = t\hat{i} + 3t^2\hat{j}$$Velocity (body starts from rest, integrate from $0$):
$$\vec{v} = \int_0^t \vec{a}\,dt = \frac{t^2}{2}\hat{i} + t^3\hat{j}$$At $t = 2$ s:
$$\vec{v} = 2\hat{i} + 8\hat{j}\ \text{m/s}, \qquad \vec{F} = 4\hat{i} + 24\hat{j}\ \text{N}$$Power:
$$P = \vec{F}\cdot\vec{v} = (4)(2) + (24)(8) = 8 + 192 = 200\ \text{W}$$Answer: 200
Solution
Data: $m = 1\ \text{g} = 10^{-3}\ \text{kg}$, $h = 1\ \text{km} = 1000\ \text{m}$, $v = 5\ \text{m/s}$.
Work-energy theorem (gravity + resistive force):
$$W_{grav} + W_{res} = \Delta KE = \tfrac{1}{2}mv^2$$Work by gravity:
$$W_{grav} = mgh = (10^{-3})(10)(1000) = 10\ \text{J}$$Change in kinetic energy:
$$\Delta KE = \tfrac{1}{2}(10^{-3})(5)^2 = 0.0125\ \text{J}$$Resistive work:
$$W_{res} = \Delta KE - W_{grav} = 0.0125 - 10 = -9.9875 \approx -9.98\ \text{J}$$Answer: D ($-9.98$)
Solution
Data: $m = 2\ \text{kg}$, $h = 10\ \text{m}$, penetration $d = 10\ \text{cm} = 0.1\ \text{m}$.
The ball starts and ends at rest, so total kinetic energy change is zero. Over the whole fall-plus-penetration, gravity acts through $(h + d)$ while the average sand force $F$ acts through $d$ opposing motion:
$$mg(h + d) - F\,d = 0$$Solve for $F$:
$$F = \frac{mg(h + d)}{d} = \frac{(2)(10)(10 + 0.1)}{0.1} = \frac{202}{0.1} = 2020\ \text{N}$$Answer: B ($2020$)
Solution
By the work-energy theorem, work done equals the change in kinetic energy:
$$W = \tfrac{1}{2}m\left(v_f^2 - v_i^2\right)$$Velocities from $v = 2x^2$:
- At $x = 0$: $v_i = 2(0)^2 = 0$
- At $x = 5$: $v_f = 2(5)^2 = 50\ \text{m/s}$
Work done:
$$W = \tfrac{1}{2}(1)\left(50^2 - 0^2\right) = \tfrac{1}{2}(2500) = 1250\ \text{J}$$Answer: C ($1250$)
Solution
Data: $m = 1\ \text{kg}$, $\theta = 30^\circ$, $v = 4\ \text{m/s}$ (constant, vertically up), $t = 2\ \text{s}$.
The block moves with the assembly at constant velocity, so its acceleration is zero and it does not slide on the incline. Static friction must balance the along-incline component of gravity:
$$f = mg\sin\theta = (1)(10)\sin 30^\circ = 5\ \text{N}$$directed up the incline to prevent the block from sliding down.
Vertical displacement in 2 s:
$$s = v\,t = 4 \times 2 = 8\ \text{m (upward)}$$The up-slope friction vector makes angle $\theta$ with the horizontal, so its vertical component is $f\sin\theta$. Work done by friction (displacement is purely vertical):
$$W_f = (f\sin\theta)\,s = (5)(\sin 30^\circ)(8) = (5)(0.5)(8) = 20\ \text{J}$$Answer: A ($20$)