Finding x for a spring using energy equations

  • Thread starter Thread starter cbasst
  • Start date Start date
  • Tags Tags
    Energy Spring
AI Thread Summary
To find how far a mass stretches a spring using energy equations, the potential energy of the mass (Ug = mgh) can be set equal to the elastic potential energy of the spring (Us = kx²/2). However, this approach yields the maximum stretch rather than the equilibrium position, where the spring force equals the weight of the mass (mg = kx). The key distinction is that at equilibrium, the mass is not in motion, while the energy method accounts for the mass's kinetic energy as it falls. Understanding this difference clarifies why the energy approach provides a different answer for x compared to the force balance method. The discussion highlights the importance of recognizing the dynamics involved when using energy equations in spring systems.
cbasst
Messages
33
Reaction score
0

Homework Statement



This isn't a homework question, just something I've been thinking about. If given a situation in which mass m is hung on a spring with constant k, I know you can set the spring force equal to the weight of the mass to find how far it stretches. Is there a way to find how far it stretches using energy equations? It seems like it should be doable, but I can't figure out how.

Homework Equations



w = mg
F = -kx
ΔUg = mgh
Us = kx2/2

The Attempt at a Solution



As I said, setting mg = kx can solve for x. How do I do it with energy though?
Ug = Us
mgx = kx2/2

But now I get a different answer for x. What did I do wrong with the energy equations?
 
Physics news on Phys.org
The energy approach will give you the farthest the mass on the spring will fall. This is not the equilibrium position. Think about it, if I let the mass drop, then when it passes the equilibrium position, it will not have zero velocity.
 
Ah. This makes sense now! Thanks!
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Spring with mass hangs from the ceiling
Replies
22
Views
1K
Time period of SHM & Energy conservation in pulley-spring-block system
Replies
24
Views
3K
Vertical spring & maximum length
Replies
6
Views
975
A block dropping onto a spring system
2
Replies
58
Views
2K
Spring Problem Involving Variables and Constants Only
Replies
3
Views
1K
Why Do Different Methods Yield Different Spring Constants in Physics Problems?
Replies
4
Views
1K
Pan suspended by a spring (Energy + SHM)
Replies
20
Views
3K
Problem of spring block system: force vs conservation of energy
Replies
20
Views
3K
Force components for mass attached to two springs
Replies
15
Views
1K
Conservation of Energy of a Spring with Weight Added
Replies
7
Views
2K

Hot Threads

Recent Insights

Back
Top