Question

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.

Answer 1

hookes law; F = -kp or some such?

Answer 2

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

Answer 3

yeah, it looks like it pretty much that:

we int it up to 95x^2

95(15)^2 - 95(7)^2 = work done

Answer 4

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

Answer 5

i get 16720 if I did it right

Answer 6

it's incorrect

Answer 7

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?