Question

Displacement, $x$ (in meters), of a body of mass $1$ kg as a function of time, $t$, on a horizontal smooth surface is given as $x = 2t^2$. The work done in the first one second by the external force is

• A. $1$ J
• B. $2$ J
• C. $4$ J
• D. $8$ J
• E. $16$ J

Hint:

When position is given as a function of time, differentiate to get velocity before applying energy relations.

Answer 1

The Correct Option is D

Concept:
Work done = change in kinetic energy: \[ W = \Delta K = \frac{1}{2}m(v^2 - u^2) \]

Step 1: Find velocity.
\[ x = 2t^2 \Rightarrow v = \frac{dx}{dt} = 4t \]

Step 2: Initial and final velocities.
At $t=0$: $u = 0$
At $t=1$: $v = 4$

Step 3: Apply work-energy theorem.
\[ W = \frac{1}{2}(1)(4^2 - 0) = \frac{1}{2} \times 16 = 8 \text{ J} \]