Solving a Simple Spring Problem: Understanding the Energy Principal

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In summary, the two principal forces in this problem are (1/2)ks2 = mgs and 2mg (upward) are the only forces acting on the weight.
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IDumb
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Hey guys, I have my physics final tomorrow and I have this simple question from a practice exam. I have gotten the correct answer by using the momentum principal, but when I try to check my work using the energy principal I get twice the correct answer. So I would appreciate it if you guys could help me figure out what I am mixing up in the energy principal. Thanks a lot.

Homework Statement



A 5 kg block is attached to a spring with stifness = 1500 N/m and relaxed length 0.2 m. You hang the spring and mass from the ceiling. When the block is hanging motionless from the spring, what is the stretch?

The Attempt at a Solution



Using the momentum principal:
Pf-Pi=0=Fnet
Fnet=Fspring-Fgrav
S=mg/Ks=.03266. (The correct answer)

Using the energy principal:
System: Block + Spring + Earth
Initial State: Block is released from rest, spring is relaxed.
Final State: Block is at rest, hanging from the spring.
*I assume that the final height of the block is 0, and the initial height is = to the stretch.* (I think this is where I am going wrong?)
Ef=Ei + Wsurroundings
Kf + Uf,spring + Uf,grav = Ki + Ui,spring + Ui,grav + 0
0 + (1/2)(Ks)(s^2) + mgh = 0 + 0 + mgh

*I assume that the final height of the block is 0, and the initial height is = to the stretch.* (I think this is where I am going wrong?)

(1/2)(1500)(s^2) = mg(s)
s=2mg/1500=.065333

Thanks again for the help
 
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Hello IDumb,

Welcome to Physics Forums!

Conservation of energy isn't the way you want to approach this problem. The disconnect here is that potential energy is not conserved in this problem.

I know it sounds a bit weird, but consider what would happen if you hold the weight in your hand while it is attached to the spring, such that the stretch is zero, and then you release the weight. The weight would fall down a little and the spring would stretch, until finally the velocity of the weight would reach zero. This is the situation where (1/2)ks2 = mgs. But it doesn't end there! The upward force on the spring is equal to 2mg (and only mg in the downward direction)! So the weight now accelerates upward, back to (almost) the point where the stretch equals 0, and the process repeats. You've created an example of simple harmonic motion.

But after awhile, due to air resistance, and other forms of friction, the system will eventually reach equilibrium at the midpoint (when the motion dies out). But by that time, much of the initial energy was lost to heat. And that's why you can't use conservation of potential energy on this particular problem. Half the energy goes into heat by the time the motion stops. :cool:
 
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  • #3
There is an analogous situation with capacitor charging.The energy stored in a charged capacitor is half of the energy needed to charge that capacitor.The rest of the energy has been lost as heat in the connecting wires.
 

FAQ: Solving a Simple Spring Problem: Understanding the Energy Principal

What is a simple spring problem?

A simple spring problem is a physics problem that involves a mass attached to a spring, where the mass is either stretched or compressed from its equilibrium position and then released. The motion of the mass is then described using equations of motion and the properties of the spring.

What is Hooke's Law and how is it related to simple spring problems?

Hooke's Law states that the force exerted by a spring is directly proportional to the amount of stretch or compression of the spring. This means that for a simple spring problem, the force exerted by the spring on the mass is directly related to the displacement of the mass from its equilibrium position.

What are the equations used to solve a simple spring problem?

The equations used to solve a simple spring problem are Newton's Second Law (F=ma) and Hooke's Law (F=-kx), where F is the force exerted by the spring, m is the mass of the object, a is the acceleration of the object, k is the spring constant, and x is the displacement of the object from its equilibrium position.

How do you determine the period and frequency of a simple spring's motion?

The period of a simple spring's motion is the time it takes for one complete cycle of motion, while the frequency is the number of cycles per second. To determine the period and frequency, you can use the equation T=2π√(m/k), where T is the period, m is the mass, and k is the spring constant.

Can a simple spring problem be solved using calculus?

Yes, a simple spring problem can be solved using calculus. The displacement, velocity, and acceleration of the mass can be described using differential equations, and these can be solved using calculus techniques such as integration and differentiation.

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