Question:

A force \( F = 5x \, \text{N} \) acts on a body and displaces it from \( x = 0 \) to \( x = 2 \, \text{m} \). The work done by the force is:

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When force is variable (like \( F = kx \)), use integration to compute work: \( W = \int F(x) \, dx \).
Updated On: Apr 19, 2025
  • \( 10 \, \text{J} \)
  • \( 20 \, \text{J} \)
  • \( 5 \, \text{J} \)
  • \( 15 \, \text{J} \)
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The Correct Option is A

Solution and Explanation

Given: - \( F = 5x \, \text{N} \) - Displacement from \( x = 0 \) to \( x = 2 \) Work done is given by: \[ W = \int_{0}^{2} F \, dx = \int_{0}^{2} 5x \, dx = 5 \int_{0}^{2} x \, dx = 5 \left[ \frac{x^2}{2} \right]_0^2 \] \[ W = 5 \cdot \frac{2^2}{2} = 5 \cdot \frac{4}{2} = 5 \cdot 2 = \boxed{10 \, \text{J}} \]
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