Work, Energy and Power JEE Main PYQs (2020–2026)

Solved Work, Energy and Power PYQs

3 recent questions with answers. The full set of 111 is free to practise.

1. JEE Main 2026MediumKinetic Energy and Work-Energy Theorem
   Q1

   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.

   • A0
   • B250
   • C1250
   • D1000

   Show answer & solution

   Correct answer: (C)

   Work done (W) is defined as the integral of force (F) over a displacement (dx):
   
   $$W={\int }_{x_i}^{x_f}Fdx$$
   
   From Newton’s Second Law, $F=ma$ , and acceleration $a=\frac{dv}{dt}$ .
   
   Using the chain rule, $a=\frac{dv}{dx}\cdot \frac{dx}{dt}=v\frac{dv}{dx}$ .
   
   $$W={\int }_{x_i}^{x_f}\left[m\left(v\frac{dv}{dx}\right)\right]dx$$
   
   $$\Rightarrow W={\int }_{v_i}^{v_f}mvdv=m{\int }_{v_i}^{v_f}vdv=m{\left[\frac{v^2}{2}\right]}_{v_i}^{v_f}$$
   
   $$\Rightarrow W=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$$
   
   $$\Rightarrow W=\Delta K.E.$$
   
   The mass of the body is $m=1kg$
   
   The velocity of body is given as function of position, $v=2x^2$
   
   Initial position is $x_i=0m$
   
   Final position is $x_f=5m$
   
   At $x_i=0$ :
   
   $$v_i=2(0{)}^2=0m/s$$
   
   At $x_f=5$ :
   
   $$v_f=2(5{)}^2=2(25)=50m/s$$
   
   On substituting these values into the Work-Energy equation :
   
   $$W=\frac{1}{2}m\left(v_f^2-v_i^2\right)$$
   
   $$\Rightarrow W=\frac{1}{2}\times 1\times \left(50^2-0^2\right)$$
   
   $$\Rightarrow W=\frac{1}{2}\times (2500)$$
   
   $$\Rightarrow W=1250J$$
   
   Therefore, the work done by the body during displacement from $x=0$ to $x=5m$ is 1250 J .
   
   Hence, the correct option is (C).
2. JEE Main 2025EasyConservative Forces, Potential Energy and Equilibrium
   Q2

   Which one of the following forces cannot be expressed in terms of potential energy?

   • AFrictional force
   • BCoulomb’s force
   • CRestoring force
   • DGravitational force

   Show answer & solution

   Correct answer: (A)

   The correct answer is:
   
   $$\text{Option A: Frictional force}$$
   
   Here’s why:
   
   Conservative vs. Non-Conservative Forces:
   
   Conservative Forces: These are forces for which the work done is independent of the path taken; they only depend on the initial and final positions. For conservative forces, you can define a potential energy function. Examples include Coulomb’s force, gravitational force, and the restoring force of a spring.
   
   Non-Conservative Forces: These forces depend on the path taken and usually result in energy dissipation (often as heat). Friction is a typical example, and hence, it cannot be described solely by a potential energy function.
   
   Example Explanations:
   
   Coulomb’s Force: It is a conservative force because the work done in moving a charge in an electric field only depends on the difference in electric potential between two points.
   
   Gravitational Force: It is also conservative. The potential energy associated with gravitational force is given by gravitational potential energy, for example, $U=mgh$ near the Earth’s surface.
   
   Restoring Force: For a spring, the force is given by Hooke’s law: $F=-kx$ and the corresponding potential energy is $U=\frac{1}{2}k{x}^2.$ This is a conservative force.
   
   Since friction is a non-conservative force that dissipates energy, it cannot be expressed in terms of potential energy.
   
   Therefore, the correct answer is Option A: Frictional force.
3. JEE Main 2025HardPower
   Q3

   An object of mass 1000 g experiences a time dependent force $\overrightarrow{F}=\left(2t\hat{i}+3t^2\hat{j}\right)N$ . The power generated by the force at time $t$ is:

   • A$\left(2t^2+18t^3\right)W$
   • B$\left(3t^3+5t^5\right)w$
   • C$\left(2t^3+3t^5\right)w$
   • D$\left(2t^2+3t^3\right)w$

   Show answer & solution

   Correct answer: (C)

   Given Information:
   
   Force: $\overrightarrow{F}=(2t\hat{i}+3t^2\hat{j})\text{N}$
   
   Mass of the object, $m=1000\text{g}=1\text{kg}$
   
   Determine Acceleration:
   
   Using Newton’s second law, $\overrightarrow{F}=m\overrightarrow{a}$ .
   
   Thus, the acceleration, $\overrightarrow{a}=2t\hat{i}+3t^2\hat{j}$ .
   
   Velocity Calculation:
   
   To find velocity, integrate acceleration with respect to time:
   
   $$\frac{d\overrightarrow{v}}{dt}=2t\hat{i}+3t^2\hat{j}$$
   
   Integrate to find $\overrightarrow{v}$ :
   
   $$\overrightarrow{v}=\int (2t\hat{i}+3t^2\hat{j})dt=t^2\hat{i}+t^3\hat{j}+\overrightarrow{C}$$
   
   (Assuming initial velocity is zero, $\overrightarrow{C}=0$ , hence $\overrightarrow{v}=t^2\hat{i}+t^3\hat{j}$ ).
   
   Calculate Power:
   
   Power is defined as the dot product of force and velocity vectors:
   
   $$P=\overrightarrow{F}\cdot \overrightarrow{v}=(2t\hat{i}+3t^2\hat{j})\cdot (t^2\hat{i}+t^3\hat{j})$$
   
   Computing the dot product gives:
   
   $$P=(2t\cdot t^2)+(3t^2\cdot t^3)=2t^3+3t^5$$
   
   The power generated at time $t$ is $P=(2t^3+3t^5)\text{W}$ .

Topic-wise question count

Where JEE Main has asked from Work, Energy and Power, most frequent first.

• Kinetic Energy and Work-Energy Theorem34
• Work Done by Constant and Variable Forces24
• Power18
• Conservative Forces, Potential Energy and Equilibrium13
• Conservation of Mechanical Energy and Vertical Circular Motion11
• Potential Energy of a Spring11

Year-wise Work, Energy and Power questions in JEE Main

Number of questions from this chapter in each year's paper(s).