How Do You Calculate the Energy of a Spring Using Physics Principles?

AI Thread Summary
To calculate the energy of a spring, the potential energy is given by the formula U = 1/2 kx^2, where k is the spring constant and x is the displacement. The discussion highlights the relationship between force and displacement, emphasizing that the area under the force vs. distance graph represents the work done on the spring. Participants express confusion over specific parts of a problem, particularly regarding the initial kinetic energy and how values like 3J were derived. The conversation clarifies that the work done on the spring is equal to the potential energy stored, and the conservation of energy principle is crucial in solving these types of problems. Understanding the integral of force over distance is essential for accurately calculating energy in spring systems.
muna580
In my class, we are curretnling doing work/engery s tuff. I am trying to figure out how to do this problem but I dont' know how. Can someone please give me a STEP by STEP explanation of how to do this? I don't understand how to do it.

http://img228.imageshack.us/img228/4813/untitled1tj3.jpg
 
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We can help you through the steps. First, state the definition of a spring constant, or at least write the equations you know that depend on the spring constant
 
F=kx Force for Spring
K=1/2kx^2 Kenetic Engery for spring
 
muna580 said:
F=kx Force for Spring
K=1/2kx^2 Kenetic Engery for spring
The first part of the graph is a graph of force versus distance. What is the slope of that portion of the graph, and how do you think that is related to the spring constant of the inner spring?

1/2kx^2 is the potential energy of the spring, not kinetic energy.
 
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I have the solutions to this problem, but there are 3 parts in the solutions which I don't understad.

I don't understand why the answer to part C is 2J.

In part D, they say,
K(inital) = 1/2mv_o^2.
3J = (1/2)(5)v_o^2.

How did they get 3J as the inital kenetig enegery?
 
muna580 said:
I have the solutions to this problem, but there are 3 parts in the solutions which I don't understad.

I don't understand why the answer to part C is 2J.

In part D, they say,
K(inital) = 1/2mv_o^2.
3J = (1/2)(5)v_o^2.

How did they get 3J as the inital kenetig enegery?
From the work done by the force that stopped the mass. C) is the work done while the mass moves from B to C. D) is the total work done.
 
Huh? I don't get it. I am so confused.
 
muna580 said:
Huh? I don't get it. I am so confused.
The reason a spring has potential energy 1/2kx^2 is because that is the amount of work it takes to stretch or compress a spring. Work is the force times the distance over which the force is applied. When the force is variable, work is found by integrating the vaiable force over the distance. If you have not had calculus, the integral is equal to the area under a force vs distance curve. The work done on the spring while the mass moves fro A to B is the area of the triangle, and from B to C it is the area of the trapezoid.

In this problem, all the kinetic energy of the mass is converted into potential energy in the springs because the mass applies a force on the springs as shown by the graph.
 
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I am curretnly taking calculus. I am still a little confused, but check this out, so far, I understand the follwoing.

My Understing.

a) What is the k? F = kx. By looking at the graphy, after traveling .10m, the force is 20N.

F = kx
20 =.1k
k = 200

b) Since the spring is losing enegery,...

(1/2)kx^2 = Energy
(1/2)(200)(.1^2) = 1J.

c) I STILL DON"T UNDERSTAND HOW TO DO IT.

d) I STILL DON"T UNDERSTAND HOW TO DO IT.

e) I STILL DON"T UNDERSTAND HOW TO DO IT.

One more thing, why is ΔK = ΔU?
 
  • #10
Correct me if I am wrong but change in Kinetic Energy is simply Work done. You can find work done by taking the area under the triangle for the first question and the area under the trapezoid for the second question. Add the two Works together to give you what was the initial KE. Since you already have the mass you can use the KE equation to find V(initial).

For part e I think you should just use the same method as you did in part a
 
  • #11
muna580 said:
One more thing, why is ΔK = ΔU?

Conservation of Energy. Since all Forces acting on the object are conservative forces (no friction, air drag, explosions that require heat or energy) all the energy remains in the system as either K or U and can only move between these two states.
 
  • #12
For part d), the solutions say

K(inital) = 1/2mv_o^2.
3J = (1/2)(5)v_o^2.

How did they get the 3J from?
 
  • #13
muna580 said:
I am curretnly taking calculus. I am still a little confused, but check this out, so far, I understand the follwoing.

My Understing.

a) What is the k? F = kx. By looking at the graphy, after traveling .10m, the force is 20N.

F = kx
20 =.1k
k = 200
This is good, but don't forget the units

b) Since the spring is losing enegery,...
The spring is gaining energy. It is doing negative work on the mass, and the mass is doing positive work on the spring. The mass is losing kinetic energy as the spring gains potential energy.

(1/2)kx^2 = Energy
(1/2)(200)(.1^2) = 1J.
Yes. And note that this is the area under the triangle. Anytime you have a graph of force vs distance in the direction of the force you can find the work by finding the area.

c) I STILL DON"T UNDERSTAND HOW TO DO IT.
The work is the area of the trapezoid between points B and C. It is also the sum of the energies stored in the two springs. The order of the questions tells me they wanted you to do this part using the area, but you could also find the spring constant for the second spring and use the compression of the first spring from A to C, plus the compression of the second spring from B to C to find the total work (total energy stored in the springs). Then find the work between B and C, by subtracting the work done between A and B. It is much easier to just find the area under the curve.

d) I STILL DON"T UNDERSTAND HOW TO DO IT.
The total work done on the springs is the total area under the curve from A to C. This is the potential energy stored in the springs when the mass comes to rest. It is also the total kinetic energy the mass had before it started slowing down. From this energy and the kinetic energy equation you can find the initial velocity.

e) I STILL DON"T UNDERSTAND HOW TO DO IT.
The steeper slope is the combined spring constant of the two springs. You can extend the first line out to x = .15 (or use the now known spring constant of the inner spring) to see how much force is being applied by the inner spring compressed .15m. The remaining force is compressing the outer spring from B to C, a distance of .05m. If you work this out you will see that the difference between the two slopes is the spring constant of the second spring.

One more thing, why is ΔK = ΔU?
Because no work is done by any disipating force like friction. The mass does work on the springs and gives them energy. The springs do work against the mass (force in opposite direction to the motion) and take energy from it.
See the annotations in blue in the quote area.
 
  • #14
OlderDan said:
See the annotations in blue in the quote area.

Wow, that was awosme, I understand everybthing. BUT, big BUT, huge BUT. lol

Umm, I understood what you said for part C, but I think finding the area under the graph is cheating. :). I mean, I like to learn things the real mathemcial way. I don't even know why we are using the area under the curve so...

Ummm, can you like write out the equation/solve part C, wihtout doing the area under the curve?

Also, I know finding the area understand the curve is taking the intergral, which is the same as the anti-dervertive, but why are we taking the area under the graphy of a F vs distance? What does it give us? How why?
 
  • #15
muna580 said:
Wow, that was awosme, I understand everybthing. BUT, big BUT, huge BUT. lol

Umm, I understood what you said for part C, but I think finding the area under the graph is cheating. :). I mean, I like to learn things the real mathemcial way. I don't even know why we are using the area under the curve so...

Ummm, can you like write out the equation/solve part C, wihtout doing the area under the curve?

Also, I know finding the area understand the curve is taking the intergral, which is the same as the anti-dervertive, but why are we taking the area under the graphy of a F vs distance? What does it give us? How why?
It is not cheating at all. The formula for the energy stored in a spring is a consequence of the fact that the integral of Fdx = kxdx is 1/2kx^2. We get the that formula because 1/2kx^2 is the area under the curve.

Between points B and C, there is an equation for the line that is the force acting on the spring. There are various ways of writing that equation, but the simplest is F = 800(N/m)x - 60N. The work required to move a tiny distance dx in the direction of an applied force is

dW = Fdx = [800(N/m)x - 60N]

This is the definition of work. Integrating as x varies from .1m to .15m gives

W = 400(N/m)(.15)^2 - 60N(.15) - [400(N/m)(.10)^2 - 60N(.10)] = 2 J

If you must use the formula, then you need to do part E first to find the spring constant of the second spring. At C the inner spring is providing 30N of force, so the outer spring is providing 30N of force. The outer spring is compressed .05m so k_2 = 30N/.05m = 600N/m. Note that the difference in slopes of the lines is 800N/m - 200N/m = 600N/m.

The total energy stored in the spring at B is

U = 1/2k_1(.1m)^2 = 100*.01 J = 1J

The total energy stored in both springs at C is

U = 1/2k_1(.15m)^2 + 1/2k_2(.05m)^2 = 100*.0225 J + 300*.0025 J = 3J
 
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  • #16
Wow, thanks a lot man. I bothered you so much. But now I understand everything. Sorry for all that trouble.
 
  • #17
It's always a good idea whenever you see a graph to look at the units on the axes and think "What will I get by dividing? (derivitive)" and "What will I get by multiplying? (integral)"

Law of Conservation of Energy is also a good thing to turn to when youre not sure about anything. My teacher told us to write LCOE=Law of Conservation of Energy at the beginning of the AP and then write LCOE APPLIES at the beginning of problems dealing with this sort of thing. Its usually 1 point on the rubrick
 

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