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Question:

The force on a particle as a function of displacement x (in the x-direction) is given by F = 10 + 0.5x. The work done corresponding to the displacement of the particle from x = 0 to x = 2 units is?

18 J

25 J

29 J

21 J

Solution:

The work done due to a small displacement dx by force F will be dW = Fdx. Integrating, W = ∫Fdx = ∫₀² (10 + 0.5x)dx = [10x + 0.5x²/2]₀² = 10(2) + 0.5(2²/2) = 20 + 1 = 21 J