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Spring Stretching Problem

  • Thread starter JoeCho
  • Start date
  • #1
2
0
1. Homework Statement

A 5 kg weight hangs on a spring with spring constant k = 0.038 n/m. How far does the spring stretch?


2. Homework Equations

Fsp = -kX

Fnet = Fsp + Fg


3. The Attempt at a Solution

So I already know that Fsp = 0.038X and that Fnet = ma = 5 (-9.8)

So -49 = 0.038X + Fg.

But isn't the force of gravity equal to 5(-9.8) as well? Am I approaching the problem incorrectly?
 

Answers and Replies

  • #2
32
0
So the spring is hanging. Its not accelerating. Its just dangling there. So sum of forces is zero. Try that.
 
  • #3
2
0
So if Fnet = 0

| Fsp
o
| Fg

Fsp + Fg = Fnet = 0

kX + mg = 0

0.038X + 5 (-9.8) = 0

0.038X = 49

X = 1289.47 Meters?

Thanks I didn't think about it not accelerating.
 
  • #4
32
0
Looks good. Didn't check numbers. Glad I could help. Good luck on your future physic endeavors.
 

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