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Homework Help: Find the spring constant and speed?

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A cube of mass m = 5 kg is released from rest and falls a height of h1 = 2 m before hitting a vertical spring which brings the cube to a stop (momentarily) after being compressed by h2 = 50 cm.


    Assume the spring has no mass, heating of the spring and the cube is negligible, and there is no air resistance.

    1. Which of the following statements are true for the system which includes Earth, the cube, and the spring: (There are more than 1 true answers)

    A. The mechanical energy of the system is preserved.
    B. The gravitational potential energy of the cube is converted into elastic potential energy of the spring at maximum compression.
    C. The kinetic energy of the cube at the time of impact is equal to the elastic potential energy of the spring at maximum compression.
    D. At no point during the motion of the cube will the net force on it be zero.

    2. Calculate the spring constant k = ____ N/m (use g = 9.81 m/s^2 and round your answer to the nearest integer).

    3. What is the speed of the cube at a height h3 = 20 cm below the point of impact?

    A. 6 m/s.
    B. 0 m/s.
    C. 21 m/s.
    D. 980 m/s.

    2. Relevant equations

    3. The attempt at a solution

    For the first question, I think it is A and B?
    I don't know about the 2nd and 3rd question. Please help
  2. jcsd
  3. Oct 11, 2009 #2


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    I believe A and B are true.

    For the 2nd part, re-read what the option of B in part 1 is saying and think about what it means in terms of equations.
  4. Oct 11, 2009 #3
    So the equation is W = mgh and W = 0.5 kx^2 ?
    So W = 5*9.81*2 = 98.1
    And W = 0.5*k*(0.5m)^2 = 0.125 k ?

    Is it like this
  5. Oct 11, 2009 #4
    How should I proceed further??
  6. Oct 11, 2009 #5


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    At the initial height is point A.

    The point at the spring in equilibrium is B (this is our reference line, so heights are positive or negative depending on where it is in relation to this point)

    The point at maximum compression is C.

    At what is the energy at point A?

    At point C, what energies is present?
  7. Oct 11, 2009 #6
    At point A, there is potential energy = mgh is there.
    At point C, potential energy of the spring is there = 0.5 k x^2 is present and I guess mgh is also present?
  8. Oct 11, 2009 #7


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    Yes and yes

    So by conservation of energy we will have mgh1=0.5kx2+mgh2

    What is h2 here knowing how we are measure the heights?
  9. Oct 11, 2009 #8
    H1 = 2m
    H2 = 0.5 m
    So then what is x? 2-0.5?
  10. Oct 11, 2009 #9


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    so mg(h1-h2)=0.5kx2

    ok we know h1=2 and x=0.5.

    So h1 to the bottom of the spring is 2m and the distance between B and C is 0.5m, so what is the distance of C to the bottom of the spring?
  11. Oct 11, 2009 #10
    How do I find that without knowing the height of the spring?
    Is it 1.5? Or 0.5?
  12. Oct 11, 2009 #11
    If you take h1 as the height of the spring, which is 2, then h2 will be 1.5
  13. Oct 11, 2009 #12


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    Ignore my last post.

    See how I said to take the reference point as point B?

    at maximum compression, in relation to point B, h2=-0.5m

    so your equation is mgh1=0.5kx2
    mgh2 and they told you x=0.5m
  14. Oct 11, 2009 #13
    What? I didn't understand, what is h2 again?
    How is h2= -0.5

    H1 = 2, x = 0.5, and the equation is mg(h1-h2) = 0.5 k x^2 right?

    I understood the H1, x and the equation part. I didn't understand h2 and how do you get it.
  15. Oct 12, 2009 #14


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    ok I probably confused you with that. Let me try to clarify.

    B (0 potential energy or reference point...h=0)

    B is the top of the spring such that AB=2m (h1)

    C is the maximum compression of the spring such that BC=0.5m

    Now at A, we know E1=mgh1.

    Meaning that relative to the point B, E1=mg(h1-0)

    Now at maximum compression, the height relative to B is h2= -0.5m since it is on the other side of the reference point.

    So the conservation of energy equation becomes mgh1=(1/2)kx2+mgh2
  16. Oct 12, 2009 #15
    I get it now.
    So k = 981 N/m?

    And for 3rd part, is it 0 m/s?
  17. Oct 12, 2009 #16


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    :uhh: that spring looks fishy

    but anyhow, for the third part it is the same process. Now instead of the right side being elastic and graviational pe, there are these two plus what type of energy?
  18. Oct 12, 2009 #17
    I don't know what you said. Can you please reply soon, I have to submit in 1 hour!
  19. Oct 12, 2009 #18


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    ok well it's like this at h=2m, there is mgh right?

    this is converted into gravitation pe at the height 0.2m (relative to B),elastic potential energy of the spring and what other type of energy? (it is moving so what energy does it have?)
  20. Oct 12, 2009 #19
  21. Oct 12, 2009 #20
    So how will be the equation?
  22. Oct 12, 2009 #21
    mgh = 1/2 k x^2 + 1/2 m v^2?
  23. Oct 12, 2009 #22
    Then what is h and x?
    h= 2m
    x = 0.2 m?
  24. Oct 12, 2009 #23
    If the values and the equation are right, then I get answer as 5.6 m/s.. Is it right?
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