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.
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?
4 answers
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
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
2 answers