a force of 4x2+3x acts on a particle which displaces the particle from x=2m and x=3m . the work done by the force is
Answers
Explanation:
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
Explanation:
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