Question

A body of mass 3 kg is under a force, which causes a displacement in it given by $S = \frac{t^3}{3}$ (in m). Find the work done by the force in first 2 seconds:

• A. 2.4 J
• B. 3.8 J
• C. 5.2 J
• D. 24 J

Hint:

$v = \frac{ds}{dt}$ and $a = \frac{dv}{dt}$. Always use Work-Energy theorem for displacement-time functions!

Answer 1

The Correct Option is D

Step 1: Concept
Work-Energy Theorem: Work done equals the change in kinetic energy ($W = \Delta KE$).

Step 2: Meaning
Displacement $S = t^3/3$. Velocity $v = \frac{dS}{dt} = t^2$.

Step 3: Analysis
At $t=0$, $v_i = 0$. At $t=2$, $v_f = (2)^2 = 4$ m/s. Work done $W = \frac{1}{2} m (v_f^2 - v_i^2) = \frac{1}{2} \times 3 \times (4^2 - 0) = \frac{3}{2} \times 16 = 24$ J.

Step 4: Conclusion
The work done is 24 Joules.

Final Answer: (D)