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Speed of a block into a spring

  1. Nov 5, 2006 #1
    A 0.252-kg block is dropped straight downwards onto a vertical spring. The spring constant of the spring is 57 N/m. The block sticks to the spring and the spring compresses 0.13 m before coming to a momentary halt. What is the speed of the block just before it hits the spring?


    well the restoring force F is equal to -kx. I need to find out how many seconds the block was in the air prior to coming in contact with the spring and compressing it .13m. This one I'm having a rough time visualizing though, any tips you have would be great....thanks again
     
  2. jcsd
  3. Nov 5, 2006 #2

    radou

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    Hint: use energy conservation to find the height from which the block is thrown. The rest is simple kinematics.
     
  4. Nov 5, 2006 #3
    PE1 + KE1 = PE2 + KE2
    .252kg*9.81m/s^2*h + .5*.252*0^2 = .252 * 9.81 * h + .5 * .252 * v^2

    I guess if I knew the height I could plug it into the equation and then figure out the value of v^2 - is this correct?
     
  5. Nov 5, 2006 #4
    * ahh - E = mv^2 /2+ kx^2 / 2

    once I find total E I can set them equal and solve for v ...
     
  6. Nov 5, 2006 #5

    radou

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    Note that the gravitational potential energy of the block in the air before its release is mg(h + 0.13). As it falls on the spring, and as the compression in the string equals 0.13, the gravitational potential energy of the block equals zero, while the spring potential energy equals 1/2*k*0.13^2.
     
  7. Nov 5, 2006 #6
    So kx^2 /2 = mgh

    57(.13^2) /2 = .252 (9.81)(h)

    h=.1948m+.13= .32483m

    PE+KE = PEf+KEf

    .252*9.81*(.32483) + .5*.252*v^2 = .252*9.81*-.13 + .5*.252*v^2

    But, if I work this one out then my v^2 cancels out leaving me with no solution.... am i doing something wrong? thanks a lot
     
  8. Nov 5, 2006 #7

    radou

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    So far, so good, except 57(.13^2) /2 = .252 (9.81)(h + 0.13). Now solve for h.
     
  9. Nov 5, 2006 #8
    57(.13^2) /2 = .252 (9.81)(h + 0.13)

    .48165 = .247 (h+.13)
    h = .0648 m

    .252 * 9.81 * .0648 + 0 = .252 * 9.81 * -.13 + .5 *.252 v^2
    .16019=-.3214+.126v^2
    v^2=3.822
    v=1.955m/s

    Hmm - this seems correct buy my book says otherwise

    * edit, yeah I'm still unsure about this one...
     
    Last edited: Nov 5, 2006
  10. Nov 5, 2006 #9
    * edit, yeah I'm still unsure about this one...
     
  11. Nov 5, 2006 #10

    radou

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    Your height is correct. Your energy conservation seems wrong. Now you can define the zero-level of potential energy as the point just before the block reaches the spring. So, you've got: mgh = 1/2 mv^2. Or, you can calculate the velocity by simply using the kinematic equation for a free fall by finding the time first, and then calculating the velocity.
     
  12. Nov 5, 2006 #11
    Ok that worked, but I'm a little confused. I thought you had to set the SUM of PE1 & KE1 equal to PE2 and KE2. Thanks
     
  13. Nov 6, 2006 #12

    radou

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    Yes, you have to set the sum of PE1 and KE1 equal to the sum of PE2 and KE2.
     
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