Question

A force \(F = (10 + 0.5x)\) N acts on a particle in the x-direction. The work done by the force in displacing the particle from \(x = 0\) to \(x = 2\) m is

• A. \(63\ \text{J}\)
• B. \(42\ \text{J}\)
• C. \(31.5\ \text{J}\)
• D. \(21\ \text{J}\)

Hint:

When force varies with position, work is always calculated using integration.

Answer 1

The Correct Option is D

Step 1: Write work done expression.
\[ W = \int_{0}^{2} (10 + 0.5x)\,dx \]
Step 2: Integrate.
\[ W = \left[10x + 0.25x^2\right]_0^2 \]
Step 3: Substitute limits.
\[ W = (20 + 1) - 0 = 21\ \text{J} \]
Step 4: Conclusion.
The work done by the force is \(21\ \text{J}\).