At a certain point, when a spring is stretched near its elastic limit, the sprin

Question

At a certain point, when a spring is stretched
near its elastic limit, the spring force satis?es the equation
F = ?? x + ? x^3,
where ? = 8 N/m and ? = 1500 N/m3.
Calculate the work done by the spring when
it is stretched from its equilibrium position to
0.1 m past its equilibrium.
Answer in units of mJ

dW = F(x) dx
W = integration of (-8x + 1500 x^3) dx
integrate and evaluate the integration from
x = 0 to x = .1 that gives

W = -8 S xdx + 1500 S x^3 dx

W = -8 x^2 + 1500^4
--- --------
2 4

Then evaluate this equation for x = .1 (the upper limit) and subtract evaluation for x = 0 (lower limit)

1.266 x 10^12

4x^2 + 375^4
4*.1^2 + 375*.1^4 =
.04 + 0.0375 = 0.0775J
into mJ........77.5mJ

I entered this into Quest and got it wrong. Not sure where my mistake is.....

Explanation / Answer

W = -integration of (-8x + 1500 x^3) dx =-(-4x^2 + 1500 x^4 /4) = 4x^2 - 375x^4 W(0.1) - W(0) = 0.04 - 0.0375 = 0.00775 = 7.75 mJ Note: we put a negative sign in 'W' equation, since force and distance are in opposite directions

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