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Energy of a spring physics

  1. Nov 2, 2006 #1
    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.

  2. jcsd
  3. Nov 2, 2006 #2


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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
  4. Nov 2, 2006 #3
    F=kx Force for Spring
    K=1/2kx^2 Kenetic Engery for spring
  5. Nov 2, 2006 #4


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    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.
    Last edited: Nov 2, 2006
  6. Nov 2, 2006 #5
    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?
  7. Nov 2, 2006 #6


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    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.
  8. Nov 2, 2006 #7
    Huh? I don't get it. I am so confused.
  9. Nov 2, 2006 #8


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    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.
    Last edited: Nov 2, 2006
  10. Nov 2, 2006 #9
    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.




    One more thing, why is ΔK = ΔU?
  11. Nov 2, 2006 #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
  12. Nov 2, 2006 #11
    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.
  13. Nov 2, 2006 #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?
  14. Nov 3, 2006 #13


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    See the annotations in blue in the quote area.
  15. Nov 3, 2006 #14
    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?
  16. Nov 3, 2006 #15


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    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
    Last edited: Nov 3, 2006
  17. Nov 3, 2006 #16
    Wow, thanks alot man. I bothered you so much. But now I understand everything. Sorry for all that trouble.
  18. Nov 3, 2006 #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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