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: Pick the easiest method.
Instead of finding the force, we use the work-energy theorem: the work done by the force equals the change in kinetic energy. \[ W = \Delta KE = \tfrac{1}{2}m v_f^2 - \tfrac{1}{2}m v_i^2 \]

Step 2: Get the velocity from displacement.
The displacement is $S = \dfrac{t^3}{3}$. Velocity is the time derivative: \[ v = \frac{dS}{dt} = t^2 \]

Step 3: Find the speeds at the two times.
At $t = 0$: $v_i = 0^2 = 0$. At $t = 2\ \text{s}$: $v_f = 2^2 = 4\ \text{m s}^{-1}$.

Step 4: Apply the theorem.
With mass $m = 3\ \text{kg}$, \[ W = \tfrac{1}{2}(3)\left(4^2 - 0^2\right) = \tfrac{1}{2}(3)(16) = 24\ \text{J} \]

Step 5: Conclusion.
The work done in the first 2 seconds is 24 J. \[ \boxed{24\ \text{J}} \]