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A force F = 20 + 10y acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is:

1. 30 J
2. 5 J
3. 25 J
4. 20 J

Correct answer: 30 J

Solution

Work done is given by the integral of force with respect to displacement: W = ∫ F dy. Substituting F = 20 + 10y and integrating from y = 0 to y = 1, we get W = ∫(20 + 10y) dy = [20y + 5y²] from 0 to 1. Evaluating, W = (20(1) + 5(1)²) - (20(0) + 5(0)²) = 20 + 5 = 30 J.

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