Hook's Law - Concept Exercise

[dbrown]

Please help as soon as possible.
Problem: Hook's law states that the distance, d, a spring is stretched varies directly with the force, F, applied to the spring.
If a force of 12 pounds stretches a spring 3 inches, how far will the spring stretch when a force of 30 pounds is applied?
I am a little confused, this is what I got so far just a formula I think, Fs = -ks or Fs = -k(x-xo) the s and o are sub.

 

[mmm4444bot]

Click HERE and scroll down the page to the Hooke's Law example of direct variation.

 

[mmm4444bot]

… F[sub:3sbpk2w0]s[/sub:3sbpk2w0] = -k[sub:3sbpk2w0]s[/sub:3sbpk2w0] …

Nope.

… or Fs = -k(x-xo)…

Okay, I missed this one on my first read.

Your second formula makes sense, and the change in the length of the spring (x - x[sub:3sbpk2w0]0[/sub:3sbpk2w0]) can be represented with a single-variable symbol.

x - x[sub:3sbpk2w0]0[/sub:3sbpk2w0] = d

Have you seen function notation?

F(d) = -k * d

The following works, too, if you prefer.

F[sub:3sbpk2w0]s[/sub:3sbpk2w0] = -k * d

(The negative sign in front of the parameter k is missing in some formulas; it all depends upon how you set up F and d, I suppose.)

Unless you're learning some finer points in physics, I would drop the negative sign, and just use positive numbers for everything: force, distance streched, and the parameter k.

F[sub:3sbpk2w0]s[/sub:3sbpk2w0] = k * d

 

[wjm11]

Hook[e]'s law states that the distance, d, a spring is stretched varies directly with the force, F, applied to the spring. If a force of 12 pounds stretches a spring 3 inches, how far will the spring stretch when a force of 30 pounds is applied?

We have a linear equation, F = kx.
We have a data point: F = 12 when x = 3.
Plug the data point into the equation to solve for k.

12 = k(3)

Once you have k, you have an equation that can be used to solve for any other F or x.