Question

A force \( F = (2 + x) \) acts on a particle in x-direction, where \(F\) is in newton and \(x\) in metre. The work done during displacement from \(x=1\) m to \(x=2\) m is:

• A. \(2 \, J\)
• B. \(3.5 \, J\)
• C. \(4.5 \, J\)
• D. None of these

Hint:

Split integrals into simple parts for faster calculation!

Answer 1

The Correct Option is B

Concept: Work done by variable force: \[ W = \int F \, dx \]

Step 1: \[ W = \int_{1}^{2} (2 + x)\, dx \]

Step 2: \[ W = \int_{1}^{2} 2 \, dx + \int_{1}^{2} x \, dx \] \[ = [2x]_1^2 + \left[\frac{x^2}{2}\right]_1^2 \]

Step 3: \[ = (4 - 2) + \left(\frac{4}{2} - \frac{1}{2}\right) = 2 + \frac{3}{2} = \frac{7}{2} \] \[ W = 3.5 \, J \]