When a certain spring is stretched beyond its proportional limit, the restoring force satisfies the equation F=-kx+βx^3. If k=8.5 N/m and β=105 N/m^3, calculate the work done by this force when the spring is stretched 0.114 m.

My attempt:

F= -(8.5)(0.114)+(105)(0.114)^3
= -0.8135 N

W= Fd = (-0.8135)(0.114) = -0.0927 J

But this is wrong. Can anyone lead me in the right direction?

work = integral of F dx

if F = -k x + b x^3
then integral of F dx is
W = -k x^2/2 + b x^4/4
evaluate that at x = 0.114
W = -[8.5 (.114)^2]/2 + 105 (.114)^3

You forgot to integrate the first and major term.

Work is the integral of F dx, not the product of F and d. Work would only be equal to F*d if the restoring force were a constant.

The integral is -(k/2)*x^2 + (â/4)*x^4,
evaluated at x = 0.114
= -0.0552 + 0.004 = -0.0548 J

W = -[8.5 (.114)^2]/2 + 105 (.114)^4

W = -[8.5 (.114)^2]/2 + (105/4) (.114)^4

To calculate the work done by the force when the spring is stretched, you need to integrate the force equation with respect to displacement.

The force equation is given as F = -kx + βx^3, where F is the restoring force, k is the spring constant, x is the displacement, and β is a constant.

To find the work done, you integrate the force equation from the initial displacement (0) to the final displacement (0.114 m) with respect to displacement.

Work (W) = ∫(F.dx)

Integrating the force equation, you get

W = ∫(-kx + βx^3)dx

To integrate each term, you use the integral rules:

∫xdx = x^2/2
∫x^3dx = x^4/4

Applying these rules, the work equation becomes:

W = -∫kx.dx + ∫βx^3.dx
= -∫kx^2/2.dx + ∫βx^4/4.dx

Integrating each term separately:

W = -k/2 * ∫x^2.dx + β/4 * ∫x^4.dx

Now substitute the limits of integration (initial and final displacement):

W = -k/2 * [x^3/3]₀ˉ˙˙.¹⁴ + β/4 * [x^5/5]₀ˉ˙˙.¹⁴
= -k/2 * [(0.114)^3/3 - (0)^3/3] + β/4 * [(0.114)^5/5 - (0)^5/5]
= -k/2 * (0.114)^3/3 + β/4 * (0.114)^5/5

Now you can plug in the given values for k and β and calculate the work done.