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Block and Constant Spring

  1. Aug 24, 2008 #1
    1. The problem statement, all variables and given/known data

    The green block is falling at a speed of 22 m/s and is 19 meters above the spring. The spring constant is 3860 N/m, to the nearest tenth of a cm how far is the spring compressed?

    The Block Weighs 2 Kg

    I have no idea on how to solve this.

    http://wps.prenhall.com/wps/media/objects/224/229574/ch5/f8-1.gif

    That is the image I am given to help with this. Any help please?
     
  2. jcsd
  3. Aug 24, 2008 #2
    Use the conservation of energy. The block will have gravitational potential energy at it's max height (19m). This will be equal to the energy that the spring uses to bring the block to a halt.

    If you haven't yet read my replies to your other thread concerning spring energy, then do so. This problem will become quite easy upon doing so.
     
  4. Aug 24, 2008 #3
    Well i Found the total mechanical Energy of the block which is 836 j. And the velocity just before it hits the spring is 28.91 m/s. So therefore it hits the spring with a force of 57.82 n. What would be the next step?
     
  5. Aug 24, 2008 #4
    Oops I misread your problem a little bit. Didn't realize that it had an initial velocity. Thus an easier approach (at least that I can think of right now) would be to gather the velocity at which it hits the spring, which is 29.3 m/s. Your calculations seem to be off a little bit. Just use:

    [tex]
    v^2 = v^2_0 + 2gy
    [/tex]

    Then, you can calculate the kinetic energy of the block right before it hits the spring, which your 836j seems to be incorrect. Then set that answer to the spring energy equation and simply solve for the amount of meters that the spring compresses. Afterwards convert the meters to centimeters as your problem states.
     
  6. Aug 24, 2008 #5
    For the velocity I achieve 24.8 m/s just before it hits the block and then i found the energy of 1211.84. If i use that in the Work= Force x Mass [1211.84= F x 2kg] I get the force correct? If that is right then the force of the block is 605.52. So then I plug that into the equation x= spring equilibrium - Force of block/Spring Constant Force. That gives me the displacement in negative meters. I convert it to centimeters and disregard the negative sign. I achieve 157 cm. Is that correct?
     
  7. Aug 24, 2008 #6
    First, explain how you got 24.8m/s and 1211.84j.
     
  8. Aug 24, 2008 #7
    I used the formula you gave me which was v^2= 22m/s^2 + 2(9.8)(19) which equals 29.26 (whoops!) is that correct? Then I put it into k=1/2 x m x v^2 ? And when that answer is found put it into the simple work equation of w=f x m. So that 856.4= f x 2. Therefore the force is 428.2 n/m.
     
  9. Aug 24, 2008 #8
    Ok, the 856.4j is correct. But just go ahead and set it equal to the spring energy equation:

    [tex]
    E =.5kx^2
    [/tex]

    Then solve for x.
     
  10. Aug 24, 2008 #9
    So the equation is 856.4= 1/2 x 3860 x X^2?

    I get the answer 66.6 cm or.6661 m

    Is this correct?!
     
  11. Aug 24, 2008 #10
    Indeed, that is correct.
     
  12. Aug 24, 2008 #11
    I'm sorry to write this but there is a second part I need help with =/

    In the previous problem to the nearest hundredth of a meter to what height will the block rise after it hits and leaves the spring?

    Would I just figure out the force that can be exerted by the spring on the object from the distance from the spring's equilibrium? Then what is next?
     
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