A body of mass 2 kg is moving along x -direction such that its displacement as function of time is given by x ( t )=α t 2 +β t +γ m , where α=1 m / s 2 , β=1 m / s and γ=1 m . The work done on the body during the time interval t =2 s to t =3 s , is \ \ \ \ J.

Physics · Work, Power & Energy

JEE Main 2026 — 21 January, Evening Shift — Question 42

A body of mass 2 kg is moving along x -direction such that its displacement as function of time is given by $\mathrm{x}(\mathrm{t})=\alpha \mathrm{t}^{2}+\beta \mathrm{t}+\gamma \mathrm{m}$ , where $\alpha=1 \mathrm{~m} / \mathrm{s}^{2}$ , $\beta=1 \mathrm{~m} / \mathrm{s}$ and $\gamma=1 \mathrm{~m}$ . The work done on the body during the time interval $\mathrm{t}=2 \mathrm{~s}$ to $\mathrm{t}=3 \mathrm{~s}$ , is $\_\_\_\_$ J.

Option A:
49
Option B:
42
Option C:
24
Correct
Option D:
12

Answer: C

Step-by-step solution

$\mathrm{x}(\mathrm{t})=\mathrm{t}^{2}+\mathrm{t}+1$ $\mathrm{v}(\mathrm{t})=2 \mathrm{t}+1$ $\mathrm{a}(\mathrm{t})=2$ $\mathrm{F}=4 \mathrm{~N}$ . Displacement $=\mathrm{x}(3)-\mathrm{x}(2)$ $=13-7=6 \mathrm{M}$ $\mathrm{W}=\mathrm{F} . \mathrm{S}=4 \times 6=24 \mathrm{~J}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Physics
Chapter: Work, Power & Energy
Topic: Kinetic Energy and Work-Energy Theorem