Compute the work required to stretch a spring from 7 to 15 cm past equilibrium, assuming that the spring constant is k=190 kg/s2.

Question

amistre64 · Answer

hookes law; F = -kp   or some such?

amistre64 · Answer

well, x = distance; so f(x) = 190x, but id have to look it up to refresh me memory

amistre64 · Answer

yeah, it looks like it pretty much that:

we int it up to 95x^2

95(15)^2 - 95(7)^2 = work done

amistre64 · Answer

or simply:  95(15^2 - 7^2)

amistre64 · Answer

i get 16720 if I did it right

anonymous · Answer

it's incorrect

amistre64 · Answer

it happens :)  i mighta misinterpreted the information given;

but the integral for the work done on a spring is:

$$\int_{7}^{15} 190x .dx$$

right?