How Far Does a Spring Stretch with a 5 kg Weight?

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A 5 kg weight hanging on a spring with a spring constant of 0.038 N/m stretches the spring based on the balance of forces. The force of gravity acting on the weight is calculated as 5 kg multiplied by -9.8 m/s², resulting in -49 N. The spring force equation, Fsp = kX, is set equal to the gravitational force, leading to the equation 0.038X + 49 = 0. Solving for X gives an incorrect stretch of 1289.47 meters, indicating a miscalculation. The discussion highlights the importance of recognizing that the system is in equilibrium, with no acceleration affecting the spring's stretch.
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Homework Statement



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


Homework Equations



Fsp = -kX

Fnet = Fsp + Fg


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?
 
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So the spring is hanging. Its not accelerating. Its just dangling there. So sum of forces is zero. Try that.
 
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.
 
Looks good. Didn't check numbers. Glad I could help. Good luck on your future physics endeavors.
 
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