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:

64 J

Explanation:

The potential energy change of the spring ∆U = -W where W = work done by force, F.

Now W = ∫F.dx

So, ∆U = - ∫F.dx = - ∫Fdxcos180 (since the spring force and extension are in opposite directions)

∆U = - ∫-Fdx

=  ∫F.dx

Since F = 40x - 6x² and x moves from x = 0 to x = 2 m, we integrate thus, ∆U =  ∫₀²F.dx

=  ∫₀²(40x - 6x²).dx

=  ∫₀²(40xdx - 6x²dx)

=  ∫₀²(40x²/2 - 6x³/3)

=  ∫₀²(20x² - 2x³)

= [20x² - 2x³]₀²

= [(20(2)² - 2(2)³) - (20(0)² - 2(0)³)

= [(20(4) - 2(8)) - (0 - 0))

= [80 - 16 - 0]

= 64 J

Answer 2

The change in potential energy will be "64 J".

Given:

• $F = 40x-6x^2$

The potential energy of spring:

→ $P.E = \int_{0}^{x}F.dx$

          $= \int_{0}^{x}(40x - 6 x^2).dx$

here,

x = 2m

then,

→ $P.E = 64 \ J$

Thus the above answer is correct.

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