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Spring: K.E. - P.E. relationship; Request shorter procedure

  1. Jun 6, 2014 #1
    Question:-
    A block of mass 10 kg moving with a velocity of 1 m/s collides with a spring of force constant 1000 N/m. Calculate the compression of the spring at the moment when kinetic energy of the block is equal to one-fourth of the elastic potential energy of the spring.

    Attempt:-
    Step 1: Determine initial Kinetic Energy of block
    Step 2: Using above value of K.E., determine maximum compression of spring using Law of Conservation of Mechanical Energy
    Step 3: Determine acceleration for the maximum compression using 3rd Kinematic equation
    Step 4: Obtain two equations with two unknowns (v & x):
    - First: v2 = u2 + 2ax (3rd Kinematic equation)
    - Second: 1/4*1/2*k*x2 = 1/2*m*v2 (from question statement)
    Step 5: Substitute v2 from First equation in the Second and solve 25x2 + 10x - 1 = 0 for x.

    Problem:-
    The above question is a 4 marks exam question which gives me 9 minutes to solve it. In addition to the 4 steps preceding it, Step 5's quadratic equation doesn't yield whole number roots. To be precise they are x=.09 & x=-.48.

    There must be a shorter procedure.

    Would appreciate advise.

    Regards,
    wirefree
     
  2. jcsd
  3. Jun 6, 2014 #2

    Simon Bridge

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    Just use conservation of energy directly. You do not need suvat equations.
    The lynchpin is that the final kinetic energy is equal to one quarter of the final potential energy.
    It's just one line.
     
  4. Jun 6, 2014 #3
    Appreciate the suggestion, Mr Bridge.

    I fail to take lead from your suggestion.

    The final K.E. is clearly zero.

    Would appreciate elaboration.

    Best,
    wirefree
     
  5. Jun 7, 2014 #4

    Simon Bridge

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    This is not what the problem statement says.
    Please reread the problem statement - it tells you the final kinetic energy is not zero when it says
    It does not say anywhere that the mass comes to rest.

    If the final kinetic energy were zero, then the calculation would be easy - all the initial kinetic energy ends up as energy stored in the spring: ##mu^2=kx^2## ... solve for x.

    If T is kinetic energy and U is potential energy, then conservation of energy says:
    $$T_i+U_i=T_f+U_f$$... you know that ##U_i=0## and you have an expression for ##T_f## in terms of ##U_f## ... so substitute and simplify.
     
    Last edited: Jun 7, 2014
  6. Jun 7, 2014 #5
    It nevertheless would when the spring sustains maximum compression. This and this alone could be termed Final K.E. (= zero).

    ImageUploadedByPhysics Forums1402120967.582200.jpg
     
  7. Jun 7, 2014 #6
    Yes the block stops at maximum compression. The question is not about maximum compression. It is about the compression that happens when potential energy is four times larger than kinetic energy. The problem is easier than you're making it out to be. Besides, your method will fail because you are using equations derived for motion at constant acceleration. That's not the case here. One more thing, the equation you're calling 3rd kinematic equation is actually called Torricelli's equation. There is no equation commonly known as 3rd Kinematic equation.
     
  8. Jun 7, 2014 #7
    As shown in the graph I attach in this post, Spring Force evolves linearly. Doesn't that indicate that the Force on the spring is constant and, therefore, so is acceleration?

    Interestingly, all 5 steps of my original post yield the same answer as the procedure outlined by Mr. Bridge in post #4.
     

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  9. Jun 7, 2014 #8
    Your OP asked for a shorter procedure.

    Initial energy = 0.5 mv^2= 5J
    Final energy = 5J = KE + PE =KE + 4KE
    KE =1J, PE =4J= 0.5 kx^2
    x = 1/5√5
     
  10. Jun 7, 2014 #9

    Simon Bridge

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    That is correct - but nowhere does it say that the spring reaches maximum compression.

    At maximum compression the final kinetic energy cannot be expressed as a fraction of the final potential energy.

    Nowhere have I said that the method you used was wrong.
    You wanted a shorter approach - this is a shorter approach.
     
    Last edited: Jun 7, 2014
  11. Jun 7, 2014 #10
    No, it does not. It shows that the force is proportional to displacement and so is the acceleration.

    Your equation (obtianed by combinig "first" and "second") should contain u and a. What values did you take for these quantities?
     
    Last edited: Jun 7, 2014
  12. Jun 8, 2014 #11

    Simon Bridge

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    Indeed - having taken another look:
    Please show your working.

    Considering that the assumption of constant acceleration is incorrect, it is a tad surprising.
    But these things happen.
     
    Last edited: Jun 8, 2014
  13. Jun 8, 2014 #12

    jtbell

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    The spring force varies with position.

    Therefore (via F = ma) the acceleration also varies with position.

    The block's position is not constant (it moves!), therefore the acceleration is not constant.

    Therefore the kinematic equations for constant acceleration do not apply.
     
  14. Jun 9, 2014 #13
    This.

    (but it is [tex]\frac{1}{5}\sqrt{2}[/tex] )
     
  15. Jun 9, 2014 #14
    Matt, that's quite cryptic. Is that your solution for x?
    :yuck:
     
  16. Jun 9, 2014 #15

    Yes, :-)
     
  17. Jun 9, 2014 #16
    I've tried and tried but I cannot get your answer. Can you show me where I went wrong?
    :cry:
     
  18. Jun 9, 2014 #17
    [tex]4 = \frac{1}{2}k x^2[/tex] So....


    ...MY GOODNESS!!! I am really sorry, mate, I read k=100 but now I see it is k=1000.

    Your result is correct. I was just reading too fast and saw 100 instead of 1000. My fault.

    Sorry :-)
     
  19. Jun 10, 2014 #18
    :smile: Thanks Matt!
     
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