Question

A force of 20 pounds stretches a spring 3/4 foot on anexercise machine. Find the work done in stretching the spring 1 foot.

Answer 1

use Hooke's law

integrate

Answer 2

ShoW me how, please. There's a reason why I'm asking

Answer 3

you know hooke's law?

Answer 4

Not well

Answer 5

\[F=-kx\]

F=20
x=3/4

Answer 6

what is k?

Answer 7

then if you are stretching from its natural length to 1 foot beyond

then the work done is

\[\int_{0}^{1}(-kx)dx\]

Answer 8

So would it be 15ft-lb?

Answer 9

no

Answer 10

20=-k3/4

\[k=-20\frac{4}{3}\]
\[k=\frac{-80}{3}\]

Answer 11

now integrate

Answer 12

Shouldn't I leave it as 80/3ft-lb

Answer 13

keep it as \[-\frac{80}{3}\]

then integrate

Answer 14

you need to calculate the work..

\[\int_{0}^{1}(-kx)dx=\int_{0}^{1}\left(-\frac{-80}{3}x\right)dx=\int_{0}^{1}\left(\frac{80}{3}x\right)dx\]

Answer 15

Final answer is supposed to be in that form. My only other option is either 40/3 or 80/9. Can't seem to get either

Answer 16

it is 40/3ft-lb

\[\int_{0}^{1}\left(\frac{80}{3}x\right)dx=\left.\frac{80}{3}\frac{x^2}{2}\right|_{0}^{1}\]

\[=\frac{40}{3}1^2-\frac{40}{3}0^2=\frac{40}{3}ft-lb\]