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Work with a spring

  1. Jan 14, 2005 #1


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    here is the problem:
    On questions a) through c), I think I got correct, here are my answer:
    a) .06 m
    spring force opposing the applied force = -3 N, and spring force = -k*d.
    So set force = -3, then solve for d, distance

    b) .18 J
    Work = F * d, we got distance from part a), and the force is given as 3 N

    c) -.09 J
    the work done by a spring when the object is stationary before and after, is given by, W = 1/2*k*x^2, so plug in the values we know, k=-50, x = .06

    Parts d) and e) are more tricky,
    the kinetic energy is maximized when the object is moving the fastest, in other words, when acceleration = 0, thus net force = 0.
    this occurs at the very end, when the spring force cancels out the applied force, but that this point, the object has come to a stop, this KE = 0.
    How do I find this?
    When originally doing this problem, I just assumed that it would reach its maximum velocty at the end (.06 m), but on closer inpection, I realized that it would have no velocity there. I solved for it anyway at that location..
    d) .06 m
    e) .09 J
    change in Ke = Word done by the applied force, plus work done by the spring, so it is .18 J - .09 J = .09 J.
  2. jcsd
  3. Jan 14, 2005 #2


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    This question is tricky, but not in the way you're thinking. Read the question more carefully. Even your answer to a) is wrong. The first part pertains to when the block stops. Will the net force on the block necessarily be zero at this point ? What can you say instead about the kinetic energy when the block stops ? Come up with an equation to relate the net work done on the block from the start to the point of the block stopping to its kinetic energy at any one time, and see what you get for answer a).
  4. Jan 15, 2005 #3


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    well obviously you think it doesnt necesarily, otherwise you wouldnt ask me the question, but I still think yes.
    if a constant force is being applied to the block, then the block will begin to accelerate...eventually the spring force will match the constant force, so then it is just mving at a constant speed, ....eventually it should reach an equalibrium between moving toward/away from the wall it is attached to and that point should be where the net force = 0, velocity = 0, wouldnt that be where the spring force = the applied force?

    well, when the block stops moving, it will have a kinetic energy of 0 J.
    when it just begins (still hasnt began to move yet), it still has a kinetic energy of 0 J. so some where in between the kinetic energy must be maximized, and at that point it has reached it highest velocity, therefiore no acceleration, therefore, net force = 0. if that is true then perhaps this is the part of the question for which the distance of .06 m is correct?
  5. Jan 15, 2005 #4
    your prt a is wrong...
    the point it stops is not its equilibium point... the spring has over strech and the block will bronce back... you have to visiulize the problem b4 pluging the number in the formulas

    when the velocity is zero, is the necessary the accelation is also zero
    Last edited: Jan 15, 2005
  6. Jan 15, 2005 #5


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    You are wrong, mrjeffy321. When the spring force = the applied force. It does not mean the block will stop moving immediately. It will continue moving.
    After that the spring force > the applied force, and the acceleration is negative. That will cause the block slowing down to a stop. Get it?
    Hope it help,
    Viet Dao,
  7. Jan 15, 2005 #6


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    Well the way I understand the problem is that the block never actually bounces all the way back, but insted just goes back and forth a couple times eventually reaching equalibrium and stoping. and then where it stops is a certain, unknown (to me) distance from the wall. but now I see that I had initially envisioned the problem incorrectly, now I can at least visualize it.

    well atleast am I correct in thinking that it's "equalibrium point" is the lcation that the block has the most kinetic energy?, and if that is true, wouldnt that location be in between (centered) the stoping place and the wall?
  8. Jan 16, 2005 #7


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    anyway, can anyone give me a very strong push in the right direction in finding these answers
  9. Jan 16, 2005 #8


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    Visualise the forces acting on the block and its velocity as a function of its displacement from the start.

    At all times there are two forces acting on the block, a constant applied force, magnitude 3 N in the positive direction (say), and a varying force due to spring tension acting in the negative direction.

    Initially, the block is at rest, its Kinetic energy (KE) is zero, and the force acting on it is +3N. It starts accelerating in the positive direction at a rate given by 3N/m where m is the mass of the block. This initial acceleration is the maximum positive acceleration.

    As the block moves in the positive x direction, the spring stretches and the tension in it increases in proportion to it (Hooke's law). As a result, the spring force acting in the negative direction gradually increases, and the acceleration in the +x direction becomes less and less the more the block moves out. Nevertheless, as long as the acceleration is positive, the block still continues to speed up.

    A point comes when the force the spring exerts on the block exactly cancels out the 3N force. Here the net force is zero, and the acceleration of the block is therefore zero. At this instant in time, the block is NOT at rest, it is still travelling. Because it has been accelerating all this time and has not yet begun decelerating this is its maximum velocity. The KE is also maximum at this point.

    The momentum of the block carries it beyond the above equilibrium point. As the block stretches the spring just a little further, you will now note that the spring force is greater in magnitude than the +3N force. Consequently, the block starts decelerating (slowing down). The block does not immediately stop and bounce back, but keeps going until the decelerating net force acting on it is enough to shave off all of its velocity and bring it to rest. When it comes to rest, the velocity is zero, KE is zero, but there is a negative force acting on the block to pull it back.

    That's the point you need to consider for the first part of the problem. It is incorrect to equate the spring force to 3N to solve for that point, because the net force is NOT zero at that point. The only think you do know is that the kinetic energy is zero at that point, and therefore the net work done on the object by the two forces acting on it is zero at that point. Can you go from there ?
    Last edited: Jan 16, 2005
  10. Jan 16, 2005 #9


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    The way i see this problem,it's not as simple as it looks.Not simple at all.Do you know to solve Cauchy problems in HS??I didn't.I learnt that in first year of college.The block is oscillating under the influence of an external force.Those are FORCED OSCILLATIONS.I never seen problems with forced oscillations in HS.
    They involve solving the following Cauchy problem:
    [tex] \frac{d^{2}x(t)}{dt^{2}}+\omega^{2}x(t)=F [/tex] (1)

    [tex] \omega=:\sqrt{\frac{k}{m}} [/tex] (2)

    [tex] x(0)=0;\frac{dx(t)}{dt}|_{t=0} =0 [/tex] (3)

    Maybe someone else will come up with a simpler solution...


    PS.The answer to "a" is 0.12m.Twice the amplitude of oscillation.
    Last edited: Jan 16, 2005
  11. Jan 16, 2005 #10


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    Further hints :

    At the point when the block stops (let that be [itex]x_1[/itex], what is the work done by the +3N force ? Use [tex]W = Fx[/tex].

    At the same point, what is the work done by the spring force ? Use [tex]W = -\frac{1}{2}kx^2[/tex].

    The sum of those two is the net work done on the block, equate to zero, and solve for [itex]x_1[/itex].

    The answers to b) and c) are equal, and they've already been calculated by the above.

    For d), you can do what you did in your original attempt, equate the spring force to 3N and solve for the force-equilibrium point (let's call that [itex]x_0[/itex].

    e) And using a similar approach to what we did for part a), you can calculate the net work done on the object at this point to give the max KE.
  12. Jan 16, 2005 #11


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    OK, here is what I have now,
    a) 1.2 m
    b) 36 J
    c) 36 J
    d) .06 m
    e) .09 J

    a) from dextercioby
    b) work from the spring at 1.2 m is equal to -.5*-50*1.2^2, that comes out to 36 J
    c) equal to part b)
    d) solve formula F=-k*d for d, where the force = -3 N and k = -50, d = .06 m
    e) work from the applied force - work from the spring (since the spring is "woking" the other way), work from the spring at .06 m is .09 J, and the work from the applied force = .18 J. that means a net work of .09 J for a kinetic energy of .09 J.

    Am I wrong on something
  13. Jan 16, 2005 #12
    Use newton's second law to find the equation of motion of the block.

    ma = F - kx

    Then use the work-kinetic energy theorem, changing the integration variable to [itex]x'[/itex].

    \Delta K & = \int_{0}^{x}(F - kx')\, dx \\
    & = Fx - \frac{1}{2}kx^{2}

    The block begins and ends at rest, so [itex]\Delta K = 0[/itex]. Solving the equation gives
    [itex]x = \frac{2F}{k} = 0.12[/itex] metres.
  14. Jan 16, 2005 #13
    Maximum kinetic energy corresponds to [itex]dK/dx = 0[/itex]. This occurs when [itex]x = F/k = 0.06[/itex] metres. K = 3*0.06 - 0.5*50*0.06^2 = 0.09 joules.
  15. Jan 16, 2005 #14


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    so then,
    a) .12 m
    b) .36 J
    c) .36 J

    I understand that explanation much better.

    and the max distance from the wall is .12 m, which is double the distance of the location where net force = 0, which is what I eluded to in post #6.
  16. Jan 16, 2005 #15


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    Yes, you got it. :smile:
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