Work Power Energy JEE Main previous year questions with solutions

5 solved JEE Main questions on Work Power Energy, free to read — no sign-in needed. The full chapter has 181 questions; sign in to attempt the remaining 176 in the exam simulator.

1. Q1JEE Main 2026 (06 Apr, Shift 2)Energy
   A body of mass $1$ kg moves along a straight line with a velocity $v = 2x^2$. The work done by the body during displacement from $x = 0$ to $5$ m is __________ J.
   1. A.$0$
   2. B.$250$
   3. C.$1250$
   4. D.$1000$
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   Answer: (C)

   Given mass $m = 1$ kg and velocity $v = 2x^2$. According to the work-energy theorem, the work done by the net force is equal to the change in kinetic energy of the body. $W = \Delta K = K_f - K_i$ At $x = 0$ m, initial velocity $v_i = 2(0)^2 = 0$ m/s. At $x = 5$ m, final velocity $v_f = 2(5)^2 = 50$ m/s. Initial kinetic energy $K_i = \dfrac{1}{2} m v_i^2 = 0$ J. Final kinetic energy $K_f = \dfrac{1}{2} m v_f^2 = \dfrac{1}{2} \times 1 \times (50)^2 = 1250$ J. Work done $W = 1250 - 0 = 1250$ J. Answer: $1250$
2. Q2JEE Main 2025 (07 Apr, Shift 1)Power
   An object of mass 1000 g experiences a time dependent force $\vec{F}=\left(2 t \hat{i}+3 t^2 \hat{j}\right) N$. The power generated by the force at time $t$ is :
   1. A.$\left(2 t^2+3 t^3\right) W$
   2. B.$\left(2 t^2+18 t^3\right) W$
   3. C.$\left(3 t^3+5 t^5\right) W$
   4. D.$\left(2 t^3+3 t^5\right) W$
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   Answer: (D)

   $\begin{aligned} & \overrightarrow{\mathrm{F}}=(2 t \hat{\mathrm{i}}+3 \mathrm{t} \hat{\mathrm{j}}) \mathrm{N} \\ & \mathrm{m}=1000 \mathrm{gm}=1 \mathrm{~kg} \\ & \overrightarrow{\mathrm{~F}}=m \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{a}}=2 \mathrm{t} \hat{\mathrm{i}}+3 \mathrm{t}^2 \hat{j} \\ & \frac{\mathrm{~d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}=2 t \hat{\mathrm{i}}+3 \mathrm{t}^2 \hat{\mathrm{j}}\end{aligned}$ $\vec{v}=t^2 \hat{i}+t^3 \hat{j}$ Power, $\mathrm{P}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{v}}$ $\begin{aligned} & P=\left(2 t \hat{i}+3 t^2 \hat{j}\right) \cdot\left(t^2 \hat{i}+t^3 \hat{j}\right) \\ & P=\left(2 t^3+3 t^5\right) W\end{aligned}$
3. Q3JEE Main 2024 (09 Apr, Shift 2)Work done
   A force $\left(3 x^2+2 x-5\right) \mathrm{N}$ displaces a body from $x=2 \mathrm{~m}$ to $x=4 \mathrm{~m}$. Work done by this force is ________ $J$.
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   Answer: 58

   $\begin{aligned} & W=\int_{x_1}^{x_2} F d x \\ & W=\int_2^4\left(3 x^2+2 x-5\right) d x \\ & W=\left[x^3+x^2-5 x\right]_2^4 \\ & W=[60-2] J=58 J\end{aligned}$
4. Q4JEE Main 2016 (10 Apr)Circular motion
   Concrete mixture is made by mixing cement, stone and sand in a rotating cylindrical drum. If the drum rotates too fast, the ingredients remain stuck to the wall of the drum and proper mixing of ingredients does not take place. The maximum rotational speed of the drum in revolutions per minute (rpm) to ensure proper mixing is close to : (Take the radius of the drum to be 1.25 m and its axle to be horizontal) :
   1. A.27.0
   2. B.0.4
   3. C.1.3
   4. D.8.0
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   Answer: (A)

   To break the circular motion max velocity at top point ${v}_{\max }=\sqrt{Rg}$ $\Rightarrow {w}_{\max }=\dfrac{{v}_{\max }}{R}=\sqrt{\dfrac{g}{R}}=\sqrt{\dfrac{10}{1.25}}$ ${w}_{\max }\left(rpm\right)=\dfrac{60}{2\pi }\sqrt{\dfrac{10}{1.25}}=27$
5. Q5JEE Main 2026 (02 Apr, Shift 2)Energy
   A spherical ball of mass $2$ kg falls from a height of $10$ m and is brought to rest after penetrating $10$ cm into sand. The average force exerted by sand on the ball is _______ N. (Take $g = 10$ m/s$^2$)
   1. A.$1980$
   2. B.$2020$
   3. C.$2000$
   4. D.$1000$
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   Answer: (B)

   Using the work-energy theorem, the total work done on the ball is equal to its change in kinetic energy. Since the ball starts from rest and finally comes to rest, the change in kinetic energy is zero. $W_{\text{gravity}} + W_{\text{sand}} = \Delta K = 0$ The work done by gravity is over the total distance $h + d$, and the work done by the resistive force of the sand $F$ is over the distance $d$. $mg(h + d) - Fd = 0$ $F = \dfrac{mg(h + d)}{d}$ Given $m = 2$ kg, $g = 10$ m/s$^2$, $h = 10$ m, and $d = 10$ cm $= 0.1$ m: $F = \dfrac{2 \times 10 \times (10 + 0.1)}{0.1}$ $F = \dfrac{20 \times 10.1}{0.1}$ $F = 20 \times 101 = 2020$ N Answer: $2020$