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Homework Help: Potential Energy & Conservative Forces #20

  1. Sep 30, 2006 #1
    Two blocks, each with a mass m = 0.348 kg, can slide without friction on a horizontal surface. Initially, block 1 is in motion with a speed v = 1.56 m/s; block 2 is at rest. When block 1 collides with block 2, a spring bumper on block 1 is compressed. Maximum compression of the spring occurs when the two blocks move with the same speed, v/2 = 0.780 m/s. If the maximum compression of the spring is 1.88 cm, what is its force constant?

    I was figuring it out using hte conservatin of energy principle, but that isnt working:

    1/2mu^2 + 1/2mu_2^2 + 1/2kx^2 = 1/2mv^2 + 1/2m_2v_2^2 + 1/2kx_2^2
     
  2. jcsd
  3. Sep 30, 2006 #2

    Doc Al

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    Why isn't it working? What values are you using for the initial/final speeds of the masses and the compression of the spring?
     
  4. Sep 30, 2006 #3
    .5(0.348)(1.56^2) + .5(.348)(0^2) + .5k(0^2) = .5(.348)(1.56^2) + .5(.348)(v/2)^2 + .5k(1.88 x 10^-2)^2

    those are all my numbers....is that right?
     
  5. Sep 30, 2006 #4

    Fermat

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    Let both masses have the same velocity after impact.
     
    Last edited: Sep 30, 2006
  6. Sep 30, 2006 #5
    i end up with a negative answer.
     
  7. Sep 30, 2006 #6

    Fermat

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    Did you use v/2 = 0.78 m/s instead of 1.56 on the rhs of the eqn ?
     
  8. Sep 30, 2006 #7
    yeah, i did, then you have to square it. I wrote my numbers up above that I was using in the equation.

    YOu end up with -0.105862 = 706.88(k)
     
  9. Sep 30, 2006 #8

    Fermat

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    .5(0.348)(1.56^2) + .5(.348)(0^2) + .5k(0^2) = .5(0.348)(1.56^2) = 0.4234464

    What do you get?
     
  10. Sep 30, 2006 #9
    .5k(0^2) = 0

    so how do I get anything when I can't even solve for k?
     
  11. Sep 30, 2006 #10

    Fermat

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    The lhs of your eqn = 0.4234464

    The rhs of your eqn = 0.217232 + 1.7672E-04 * k

    Solve for k.

    Check your arithmetic.
     
  12. Sep 30, 2006 #11
    i get

    0.423 = 0.529308 + 1.7672 x 10^-4

    Which gives a negative answer.


    Using what you wrote above I get k = 1166.9 and that answer was not correct.
     
  13. Sep 30, 2006 #12

    Fermat

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    My mistake: 0.217232 + 1.7672E-04 * k should be 0.2117232 + 1.7672E-04 * k

    Which gives k = 1.2 KN/m

    You had 0.529308 on the rhs. You will get that if you used v = 1.56 instead of v = 0.78.
    Both masses have the same speed, 0.78 m/s.
     
  14. Sep 30, 2006 #13

    Doc Al

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    No. As Fermat has already pointed out, both masses move with the same speed (0.780 m/s) when the spring is fully compressed.
     
  15. Sep 30, 2006 #14
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