Question

To stretch a certain nonlinear spring by an amount x requires a force F given by F = 40 x − 6 x 2 , where F is in Newtons and x is in meters. What is the change in potential energy ∆U when the spring is stretched 2 meters from its equilibrium position?

Answer 1

Answer:

90 J

Explanation:

Change in potential energy when the spring is stretched

ΔU = ∫[(F) dx]................... Equation 1

Where F = 40x-6x²

Substitute into equation 1

ΔU = ∫[40x-6x²]

ΔU = [40x²/2 + 6x³/3 +C]

ΔU = [20x²+2x³ +C]

If the spring is stretch from x = 0m to x = 2m

ΔU = [20×2²+2×2³+C]-[20×0²+2×0³+C]

ΔU = [80+16+C]-[C]

ΔU = 90 J.

Hence the change in potential energy of the spring = 90 J