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.

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Work, Power & Energy

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

t = 3 s

, is
____ J.

(A)49
(B)42
(C)24✓ Correct
(D)12

Answer: (C)

Solution

Given

x (t) = t^{2} + t + 1

, the velocity is

v (t) = \frac{d x}{d t} = 2 t + 1

and acceleration is

a = \frac{d v}{d t} = 2

m/s². The force is

F = ma = 2 \times 2 = 4

N. The displacement from t=2s to t=3s is

Δ x = x (3) - x (2) = (9 + 3 + 1) - (4 + 2 + 1) = 13 - 7 = 6

m. Work done:

W = F \cdot Δ x = 4 \times 6 = 24

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option

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