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A position dependent force, F = (7 − 2x + 3x²) N acts on a small body of mass 2 kg and displaces it from x = 0 to x = 5 m. Work done in joule is

1. 35
2. 70
3. 135
4. 270

Correct answer: 135

Solution

The work done by a variable force is calculated using the integral of F(x) with respect to x over the given limits. Here, W = ∫(7 − 2x + 3x²) dx from 0 to 5. Solving, W = [7x − x² + x³] from 0 to 5 = (7(5) − (5)² + (5)³) − (7(0) − (0)² + (0)³) = 35 − 25 + 125 = 135 J.

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