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Question:

When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx². where a and b are constants. The work done in stretching the unstretched rubber band by L is

aL²/2 + bL³/3

1/2(aL²/2 + bL³/3)

aL + bL²

1/2(aL + bL²)

Solution:

The work done in stretching the rubber band is given by the integral of the force over the distance stretched. The force is given by F = ax + bx².

The work done (W) is:

W = ∫₀ˡ F dx = ∫₀ˡ (ax + bx²) dx

Integrating, we get:

W = [ax²/2 + bx³/3]₀ˡ

W = (aL²/2 + bL³/3) - (a(0)²/2 + b(0)³/3)

W = aL²/2 + bL³/3