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 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:
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Split integrals into simple parts for faster calculation!
Updated On: Apr 17, 2026
- \(2 \, J\)
- \(3.5 \, J\)
- \(4.5 \, J\)
- None of these
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The Correct Option is B
Solution and Explanation
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: \[ = (4 - 2) + \left(\frac{4}{2} - \frac{1}{2}\right) = 2 + \frac{3}{2} = \frac{7}{2} \] \[ W = 3.5 \, J \]
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: \[ = (4 - 2) + \left(\frac{4}{2} - \frac{1}{2}\right) = 2 + \frac{3}{2} = \frac{7}{2} \] \[ W = 3.5 \, J \]
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