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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

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5 J

25 J

30 J

20 J

answer is C.

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Detailed Solution

$W = \int F . dS = \int_{0}^{1} (20 + 10 y) dy = 25 J$

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