Spring deflection problem using energy equations

  • Thread starter ago01
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  • #1
ago01
46
8
Homework Statement:
Given a box of mass 0.8kg how far is the spring compressed if the block is placed on the spring and does not oscillate? In other words, what is the compression of the spring at the equilibrium point of the block- spring system? The spring constant is 22 N/m.
Relevant Equations:
mg, -kx
At the equilibrium position all forces are equal to 0 (they balance). So, naturally we:

mg - kd = 0

d = mg/k

d = 0.356 m

This is the correct answer, I believe. But I want to solve it using the energy equations because I am really trying to understand energy's connection to the rest of kinematics.

At the equilibrium position all forces are 0. But there is potential energy stored in the spring and the box, and to be at equilibrium these both must cancel. Let `d` be the deflection and the ground be the 0 potential plane. Place the bottom of the spring on the 0 potential plane.

So,

mgd = (1/2)kd^2

d = (2mg)/k

d= 0.712m

Which of course is a factor of 2 larger than the actual answer. I am confused, as I am lead to believe this would be solvable both ways (one using forces, one using energy). I have asked this to several people and they seem to immediately go to the d = mg/k solution, which is fine.

Is this solvable using conservation of energy or am I spinning my wheels for no reason?
 

Answers and Replies

  • #2
Henryk
Gold Member
267
105
I can see that your energy reference point is: the spring is fully extended and the weight is placed on top of that.
The energy the will have three components: potential energy due to gravity equal -mgd, plus spring potential energy, equal to ##\frac 1 2 kd^2## and the kinetic energy ##\frac 1/2 m \dot d ^2##. Physically, when you start by placing the weight on top of the spring, you will get oscillation. The problem states, what is the position of the box at equilibrium, i.e. after the kinetic energy of the oscillations have dissipated.
At that time, the spring compression energy does not equal to the potential energy due to gravity.
 
  • #3
ago01
46
8
Thank you for your reply. I think I see now why this doesn't work. To be clear, the reason is that at position d the mass is still off the ground, so it still has gravity pulling it back down towards earth. At the instant it is at d (equilibrium) all forces cancel. But this doesn't imply that gravity suddenly turns off. "Letting the simulation run" past d would show that the weight doesn't stop at that point, but rather keeps falling beyond the equilibrium point due to gravity. So the equilibrium point is just the place where, at that instant, all forces are balanced.

I think I get it. Thank you again.
 

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