Question

A particle moves under a force \(F = 3x^2\) N. The work done by the force on the particle in displacing it from \(x=0\) to \(x=2\text{ m}\) is}

• A. 12 J
• B. 3 J
• C. 6 J
• D. 8 J
• E. 15 J

Hint:

For variable force: \[ W=\int F\,dx \] Always use integration, not \(F\times s\), when force depends on position.

Answer 1

The Correct Option is D

Work done by a variable force is: \[ W=\int_{x_1}^{x_2} F\,dx \] Here, \[ F=3x^2 \] So, \[ W=\int_0^2 3x^2\,dx \] \[ W=3\int_0^2 x^2\,dx \] \[ W=3\left[\frac{x^3}{3}\right]_0^2 \] \[ W=\left[x^3\right]_0^2 \] \[ W=2^3-0=8 \] So the work done is: \[ \boxed{8\text{ J}} \]
Hence, the correct answer is: \[ \boxed{(D)} \]