Physics · Work, Power & Energy

JEE Main 2026 — 6 April, Evening Shift — Question 3

A body of mass $1\mathrm{kg}$ moves along a straight line with a velocity $v = 2x^2$ . The work done by the body during displacement from $x = 0$ to $5\mathrm{m}$ is J.

Option A:
0
Option B:
250
Option C:
1250
Correct
Option D:
1000

Answer: C

Step-by-step solution

Work done = change in KE = $\frac12 m (v_f^2 - v_i^2)$ . At x=0, v=0 at x=5, v=225=50 m/s. KE = 0.51*2500 = 1250 J.

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Physics
Chapter: Work, Power & Energy
Topic: Kinetic Energy and Work-Energy Theorem