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Spring/Velocity problem

  1. Feb 23, 2009 #1
    1. The problem statement, all variables and given/known data
    At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large, compressed spring. The spring with a force constant k = 3600 N/m and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 75.0 kg are pushed against the other end, compressing the spring 0.385 m. The sled is then released with zero initial velocity.

    A. What is the sled's speed when the spring returns to its uncompressed length?
    B. What is the sled's speed when the spring is still compressed 0.215 m?

    2. Relevant equations
    W=K2-K1
    W=1/2kx^2-1/2kx^2
    W=1/2kx^2

    3. The attempt at a solution
    I solved A by using W=1/2kx^2 to find the work done by the spring to be 266.81J. I then used that along with V1=0, k=3600 and m=75.0 to find V2 which was 2.67m/s. I got this part right.

    Part B is where I am having some trouble.

    My first try, I tried to do the same thing as above.
    I found the work done by the spring at .210m:
    W=1/2(3600)(.210)^2=83.205J

    and then I used this in the W=K2-K1 with V1 being zero
    83.205=1/2(75)(V2)^2-1/2(75)(0)^2

    V2=1.49m/s <--- this answer was wrong.

    My second try, I did the same thing except I used W=1/2kx^2-1/2kx^2, using X1=.210m and x2=.385m. With this, I found the work done to be W=183.6J

    Using that, I plugged it into the W=KE2-KE1 to find V1:
    183.6=1/2(75.0)(2.67)^2-1/2(75.0)(V1)
    V1=1.50

    Can someone show me what I am doing wrong?
     
  2. jcsd
  3. Feb 23, 2009 #2

    LowlyPion

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    For Part B, is it .210m or .215m as in the original question you posted?
     
  4. Feb 23, 2009 #3

    rl.bhat

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    B. What is the sled's speed when the spring is still compressed 0.215 m?
    Still compressed means, is it in addition to 0.385 m?
     
  5. Feb 23, 2009 #4

    LowlyPion

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    I think the OP has understood it correctly insofar as it wants to know what the speed is as it passes .215m going to 0 detent.
     
  6. Feb 23, 2009 #5
    It's .215m, my bad.
     
  7. Feb 23, 2009 #6
    Does anyone know what I am doing wrong?
     
  8. Feb 24, 2009 #7

    LowlyPion

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    Spring Potential E(.385) - Spring Potential E(.215) = 1/2*m*v2

    3600/2*((.385)2 - (.215)2) = 1/2*75*V2

    V2 = 1800*(.102)*2/75

    V = 2.12
     
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