# Spring deflection problem using energy equations

• ago01
In summary, the equilibrium position for a spring and a box with mass on it is where all the forces are 0.
ago01
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?

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.

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.

## 1. What is the spring deflection problem using energy equations?

The spring deflection problem using energy equations is a mathematical approach to determining the amount of deflection or displacement of a spring when a force is applied to it. It involves using the principles of energy conservation to calculate the potential and kinetic energy of the spring at different points of deflection.

## 2. How do you solve the spring deflection problem using energy equations?

To solve the spring deflection problem using energy equations, you need to first identify the known variables such as the spring constant, applied force, and initial and final positions of the spring. Then, use the equations for potential and kinetic energy to set up an equation and solve for the unknown variable, which is usually the deflection of the spring.

## 3. What are the assumptions made in the spring deflection problem using energy equations?

The assumptions made in the spring deflection problem using energy equations include: the spring is ideal and obeys Hooke's Law, there is no friction or other external forces acting on the spring, and the spring is massless.

## 4. Can the spring deflection problem using energy equations be used for non-ideal springs?

No, the spring deflection problem using energy equations is only applicable for ideal springs that obey Hooke's Law. For non-ideal springs, other methods such as finite element analysis may be used.

## 5. What are some real-life applications of the spring deflection problem using energy equations?

The spring deflection problem using energy equations has various real-life applications, such as in the design of suspension systems in cars, determining the deflection of springs in mechanical devices, and calculating the stress and strain on springs in engineering projects.

• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
13
Views
259
• Introductory Physics Homework Help
Replies
14
Views
337
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
24
Views
1K
• Introductory Physics Homework Help
Replies
22
Views
470
• Introductory Physics Homework Help
Replies
5
Views
822
• Introductory Physics Homework Help
Replies
3
Views
360
• Introductory Physics Homework Help
Replies
29
Views
908
• Introductory Physics Homework Help
Replies
5
Views
345