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 143 questions; sign in to attempt the remaining 138 in the exam simulator.

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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 ________ $J$.
   Show answer & solution

   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 2023 (06 Apr, Shift 1)Non-uniform Motion
   A particle of mass $10g$moves in a straight line with retardation $2x,$ where $x$ is the displacement in $SI$ units. Its loss of kinetic energy for above displacement is ${\left(\dfrac{10}{x}\right)}^{-n}J.$ The value of $n$ will be $\underline{\phantom{0000}}.$
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   Answer: 2

   Acceleration can be written as, $a=v\dfrac{dv}{dx}$. Given: $a=-2x$(-ve sign is due to retardation) Therefore, $v\dfrac{dv}{dx}=-2x \Rightarrow vdv=\left(-2x\right)dx$ Integrating both sides, we get $\dfrac{1}{2}\left({v}_{f}^{2}-{v}_{i}^{2}\right)=-2\left(\dfrac{{x}^{2}}{2}\right)={x}^{2} \Rightarrow \left|\dfrac{1}{2}m\left({v}_{f}^{2}-{v}_{i}^{2}\right)\right|=m{x}^{2}$ $\Rightarrow |\Delta KE|=0.01{x}^{2}={\left(\dfrac{10}{x}\right)}^{-2}$ Hence, $n=2$.
5. Q5JEE Main 2021 (20 Jul, Shift 2)Rest and Motion
   A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time $t$ is proportional to:
   1. A.${t}^{\dfrac{3}{2}}$
   2. B.${t}^{\dfrac{1}{2}}$
   3. C.${t}^{\dfrac{1}{4}}$
   4. D.${t}^{\dfrac{3}{4}}$
   Show answer & solution

   Answer: (A)

   $P=$ constant $\dfrac{1}{2}m{v}^{2}=Pt$ $\Rightarrow v\propto \sqrt{t}$ $\dfrac{dx}{dt}=C\sqrt{t}$ $C=$ constant by integration, $x=C\dfrac{{t}^{\dfrac{1}{2}+1}}{\dfrac{1}{2}+1}$ $x\propto {t}^{3/2}$