Physics Q29 – JEE-MAIN 2026

A body of mass

1

kg moves along a straight line with a velocity

v = 2 x^{2}

. The work done by the body during displacement from

x = 0

5

m is _______ J.

The work done by the net force on a body is equal to the change in its kinetic energy.

Step 1: Apply the Work-Energy Theorem✦ Active

The work done ( $W$ ) by the body is equal to the change in its kinetic energy ( $Δ K$ ). This is given by the formula:

W = Δ K = K_{f} - K_{i}

Where $K_{f}$ is the final kinetic energy and $K_{i}$ is the initial kinetic energy. The kinetic energy is given by $K = \frac{1}{2} m v^{2}$ .

Step 2: Calculate Initial and Final Kinetic Energies○ Expand

Given mass $m = 1$ kg and velocity $v = 2 x^{2}$ .

At the initial position $x_{i} = 0$ m:

v_{i} = 2 (0)^{2} = 0 m/s

K_{i} = \frac{1}{2} m v_{i}^{2} = \frac{1}{2} (1) (0)^{2} = 0 J

At the final position $x_{f} = 5$ m:

v_{f} = 2 (5)^{2} = 2 (25) = 50 m/s

K_{f} = \frac{1}{2} m v_{f}^{2} = \frac{1}{2} (1) (50)^{2} = \frac{1}{2} (2500) = 1250 J

💡 Teacher's Secret Hint

Ensure correct substitution of x into the velocity equation.

Step 3: Calculate the Total Work Done○ Expand

Using the Work-Energy Theorem with the calculated kinetic energies:

W = K_{f} - K_{i} = 1250 J - 0 J = 1250 J

The work done by the body is $1250$ J.

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Physics Question 29 – JEE-MAIN 2026