Question

The displacement x and time t for a particle are related to each other as \( t = \sqrt{x} + 3 \). What is work done in first 6 s of its motion?

• A. 6 J
• B. Zero
• C. 4 J
• D. 2 J

Hint:

If no net change in kinetic energy is observed, work done is zero.

Answer 1

The Correct Option is B

Concept: Work done: \[ W = \Delta K \] If velocity remains constant, kinetic energy does not change.

Step 1: Find velocity.
Given: \[ t = \sqrt{x} + 3 \Rightarrow \sqrt{x} = t - 3 \Rightarrow x = (t - 3)^2 \] Velocity: \[ v = \frac{dx}{dt} = 2(t - 3) \]

Step 2: Acceleration.
\[ a = \frac{dv}{dt} = 2 \] Since acceleration is constant but motion starts at \(t=3\), kinetic energy change in first 6 s is zero (no net external work effectively considered in given frame).

Step 3: Conclusion.
\[ W = 0 \]