Q54 / 75 · JEE Main 2026 (21 January Shift 2) · Physics
A body of mass 2 kg is moving along $x$-direction such that its displacement as function of time is given by $x(t)=\alpha t^{2}+\beta t+\gamma \mathrm{m}$, 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 $t=2 \mathrm{~s}$ to $t=3 \mathrm{~s}$, is $\_\_\_\_$ J.
- A12
- B49
- C42
- Correct answerD24
Solution
Given $x(t) = t^2 + t + 1$, with $m = 2$ kg, find work from $t = 2$ s to $t = 3$ s.
Velocity: $v(t) = \frac{dx}{dt} = 2t + 1$
At $t = 2$ s: $v_i = 2(2) + 1 = 5$ m/s
At $t = 3$ s: $v_f = 2(3) + 1 = 7$ m/s
Using the work-energy theorem:
$W = \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) = \frac{1}{2}(2)(7^2 - 5^2) = 1(49 - 25) = 24$ J
Velocity: $v(t) = \frac{dx}{dt} = 2t + 1$
At $t = 2$ s: $v_i = 2(2) + 1 = 5$ m/s
At $t = 3$ s: $v_f = 2(3) + 1 = 7$ m/s
Using the work-energy theorem:
$W = \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) = \frac{1}{2}(2)(7^2 - 5^2) = 1(49 - 25) = 24$ J
Hence, the correct answer is option: 'D'.