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

Question

zarkon · Answer

use Hooke's law

integrate

anonymous · Answer

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

zarkon · Answer

you know hooke's law?

anonymous · Answer

Not well

zarkon · Answer

$$F=-kx$$

F=20
x=3/4

zarkon · Answer

what is k?

zarkon · Answer

then if you are stretching from its natural length to 1 foot beyond

then the work done is

$$\int_{0}^{1}(-kx)dx$$

anonymous · Answer

So would it be 15ft-lb?

zarkon · Answer

no

zarkon · Answer

20=-k3/4

$$k=-20\frac{4}{3}$$
$$k=\frac{-80}{3}$$

zarkon · Answer

now integrate

anonymous · Answer

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

zarkon · Answer

keep it as $$-\frac{80}{3}$$

then integrate

zarkon · Answer

you need to calculate the work..

$$\int_{0}^{1}(-kx)dx=\int_{0}^{1}\left(-\frac{-80}{3}x\right)dx=\int_{0}^{1}\left(\frac{80}{3}x\right)dx$$

anonymous · Answer

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

zarkon · Answer

it is 40/3ft-lb

$$\int_{0}^{1}\left(\frac{80}{3}x\right)dx=\left.\frac{80}{3}\frac{x^2}{2}\right|_{0}^{1}$$

$$=\frac{40}{3}1^2-\frac{40}{3}0^2=\frac{40}{3}ft-lb$$