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

Given: m = 2 kg and x(t) = t^2 + t + 1. Find: Work done from t = 2 s to t = 3 s. Using the Work-Energy Theorem, the work done by the net force equals the change in kinetic energy. Velocity is obtained by differentiating displacement with respect to time: v(t) = dx(t)dt = ddt(t^2 + t + 1) = 2t + 1 At ti = 2 s: vi = 2(2) + 1 = 5 m/s At tf = 3 s: vf = 2(3) + 1 = 7 m/s Now apply the work-energy theorem: W = K = 12m (vf^2 - vi^2) Substituting m = 2 kg, vf = 7 m/s and vi = 5 m/s: W = 12 2 (7^2 - 5^2) W = 49 - 25 = 24 J Therefore, the work done is 24 J. The correct option is B.

A body of mass 2 kg is moving along x-direction such that its dis… — Solution | JEE Physics

MCQEasyJEE 2026Kinetic Energy & Work-Energy Theorem

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

A
$42$
B
$24$
C
$12$
D
$49$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $m = 2 \, \text{kg}$ and $x(t) = t^2 + t + 1$ .

Find: Work done from $t = 2 \, \text{s}$ to $t = 3 \, \text{s}$ .

Using the Work-Energy Theorem, the work done by the net force equals the change in kinetic energy.

Velocity is obtained by differentiating displacement with respect to time:

v(t) = \frac{dx(t)}{dt} = \frac{d}{dt}(t^2 + t + 1) = 2t + 1

At $t_i = 2 \, \text{s}$ :

v_i = 2(2) + 1 = 5 \, \text{m/s}

At $t_f = 3 \, \text{s}$ :

v_f = 2(3) + 1 = 7 \, \text{m/s}

Now apply the work-energy theorem:

W = \Delta K = \frac{1}{2}m\left(v_f^2 - v_i^2\right)

Substituting $m = 2 \, \text{kg}$ , $v_f = 7 \, \text{m/s}$ and $v_i = 5 \, \text{m/s}$ :

W = \frac{1}{2} \cdot 2 \cdot (7^2 - 5^2)

W = 49 - 25 = 24 \, \text{J}

Therefore, the work done is $24 \, \text{J}$ . The correct option is B.

Common mistakes

Differentiating $x(t)$ incorrectly. The velocity is $v(t) = \frac{dx}{dt} = 2t + 1$ , not $t^2 + 1$ or any other expression. Always take the time derivative before using energy relations.
Using displacement difference instead of change in kinetic energy. Work done by the net force here is found most directly from $W = \Delta K$ , not from substituting $\Delta x$ into an unrelated formula.
Forgetting to square the velocities in kinetic energy. The theorem uses $\frac{1}{2}mv^2$ , so the change is based on $v_f^2 - v_i^2$ , not $v_f - v_i$ .

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Answer

Step-by-step solution

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