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Conservation of energy using a spring

  1. Oct 23, 2012 #1
    1. The problem statement, all variables and given/known data

    A 785-gram metal block is placed on a level table that is 1.6 m high. It is connected to a horizontal spring whose force constant is 3000 N/m. The block is pushed against the spring, compressing it by 0.100 m, and released. It slides along the table and goes off of the edge.

    Coefficient of kinetic friction is 0.200 (Uk)

    If the speed of the block just before it strikes the ground is 7.3 m/s, what distance did the block slide along the table?


    2. Relevant equations

    Work = delta K + delta U
    U spring= (1/2)Kx^2
    K block = (1/2)mv^2





    3. The attempt at a solution

    What I can come up with so far is that the potential energy of the spring (U) is equal to the kinetic energy done on the block (K). So in that case I set the two equations equal to eachother

    (1/2)Kx^2 = (Uk)mg(delta X) and so

    15 J = 1.54(delta X)

    so delta X = 9.74 m.


    This doesnt seem very plausable to me with such a high spring coefficient and neglecting to use the height of the table and final velocity.

    Any ideas?
     
  2. jcsd
  3. Oct 23, 2012 #2

    PhanthomJay

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    Start backwards. Use conservation of energy between the instant it leaves the table to the instant it hits the ground to solve for its speed as it left the table. Then use your work energy equation between the start point and the point it left the table.
     
  4. Oct 23, 2012 #3
    You're assuming that the block goes only on the table. You need to take into account its total energy when compressed and its total energy when hitting the ground.
     
    Last edited: Oct 23, 2012
  5. Oct 24, 2012 #4
    Okay so I think im heading in the right direction now.

    IVe got

    Work = (delta K) + ( delta Uspring) + (delta Ugravity)

    delta K = (1/2)Mvf^2 - (1/2)Mvi^2 = 20.92 J
    delta Uspring = (1/2)Kxf^2 - (1/2)Kxi^2 = -15 J
    delta Ugravity = (1/2)mghf - (1/2)kxf^2 = -12.31 J

    adding theses gives me W = -6.39

    I can also say that Wother = (coefficient of kinetic friction)(force of kinetic friction)(delta X)

    in this case how am I coming up with the force of kinetic friction? because isnt that (coefficient of kinetic friction)*(normal force)? This just seems redundant to me.

    any ideas?
     
  6. Oct 24, 2012 #5
    Nevermind! haha I ended up getting it. I was just thinking about the force of friction wrong. Somehow I neglected to include gravity in my first calculation which gave an extremely long distance. thanks for the input guys!
     
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