Question

a force of (4x^2 + 3x ) N acts on a particle which displaces it from x=2 to x=3 m .the work done by the force is

Answer 1

Since the force is variable, let's consider the small work done (dW) by the force in displacing the particle by a very small distance (dx)

dW = F•dx = (4x² + 3x)dx

For the total work done, we need to integrate both sides,

W = ∫ dW = ∫ (4x² + 3x) dx

W = [4x³/3 + 3x²/2]

Since, we need W from x = 2 to x = 3, our lower limit will be 2 and upper limit will be 3

W = [4x³/3 + 3x²/2]₂³

W = 4(3)³/3 + 3(3)²/2 - 4(2)³/3 - 3(2)²/2

W = 36 + 13.5 - 10.6 - 6

W = 49.5 - 16.6 = 32.9J ≈ 33J
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Answer 2

Answer:

dW=F.dx=(4x^2+3x)dx

integrating with limits we get

W= 32.8J