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Incline with a Spring and Loop the Loop

  1. Oct 10, 2007 #1
    A 3.0 kg block starts at rest and slides a distance d down a frictionless 32.0° incline, where it runs into a spring. The block slides an additional 24.0 cm before it is brought to rest momentarily by compressing the spring, whose spring constant is 439 N/m. What is the value of d?

    img:http://i242.photobucket.com/albums/ff106/jtdla/prob01a.gif

    I'm not sure how to start this. I'm not sure how to calculate potential/kinetic energy (which was suppose to be the theme of the homework). I'm aware that this problem uses Hooke's law, but I'm not sure how to put it all together.


    A small block of mass m = 2.0 kg slides, without friction, along the loop-the-loop track shown. The block starts from the point P a distance h = 53.0 m above the bottom of the loop of radius R = 20.0 m. What is the kinetic energy of the mass at the point A on the loop?

    img:http://i242.photobucket.com/albums/ff106/jtdla/prob17a.gif

    Outside of the potential energy, I'm not sure how kinetic energy works in a circle. Or it its no different. I'm not sure if this assumption is correct, shouldn't the PE at the top of the track equal the KE at the bottom of the loop. Would that KE then equal the PE for the look itself. I'm not sure at all how to do these problems. This conservation of energy has been really tricky for me.
     
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  3. Oct 10, 2007 #2

    learningphysics

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    In the first problem, use gravitational potential energy and elastic potential energy... at the top it has only gravitational potential energy... at the bottom it has only elastic potential energy... use conservation of energy.

    In the second problem, it doesn't matter that it is a circle or a loop... just use conservation of energy... use gravitational potential energy and kinetic energy... again look at the beginning and at the end... at the beginning it has only gravitational potential energy... at the end it has gravitational potential energy and kinetic energy (there's no need to think about the bottom of the loop).
     
  4. Oct 10, 2007 #3
    mgh= 1/2mv^2

    how would I calculate the height for potential energy?

    for the second problem:

    I thought it would only have kinetic energy at the end. Would I just plug everything into a mgh+0=0+1/2mv^2
     
  5. Oct 10, 2007 #4
    Would an equation for the loop the loop problem be
    2*9.8*53=1/2*2*v^2=32.2304

    or would it equal PE=KE+PE(at top of loop)
     
  6. Oct 10, 2007 #5

    learningphysics

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    yes, because it has both KE and PE at the top of the loop...
     
  7. Oct 10, 2007 #6

    learningphysics

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    do you mean mgh = (1/2)kx^2? use trig for height. distance = d + 0.24
    so what is the height?
     
  8. Oct 10, 2007 #7
    For the loop the loop, would the velocity be 25.432 m/s, would that also be the kinetic energy at A.
     
  9. Oct 10, 2007 #8

    learningphysics

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    how'd you get that?
     
  10. Oct 10, 2007 #9
    I set up that equation

    PE=KE+PE


    1038.8(Start)=1/2*2*v^2+392(point A) v=25.4323
     
    Last edited: Oct 10, 2007
  11. Oct 10, 2007 #10

    learningphysics

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    Height at the top of the loop is 2*20 = 40m
     
  12. Oct 10, 2007 #11
    15.9625, I forgot about 2r

    Is that the answer to the loop the loop
     
  13. Oct 10, 2007 #12
    254.801 for the kinetic energy
     
  14. Oct 10, 2007 #13

    learningphysics

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    yeah, looks right.
     
  15. Oct 10, 2007 #14
    okay, the spring one is really throwing me off

    so the distance traveled is d+.24, how can I use trig
     
  16. Oct 10, 2007 #15

    learningphysics

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    you have a right triangle... angle 32. hypoteneuse d+0.24. What's the height?
     
  17. Oct 10, 2007 #16
    could i set something up like cos^-1(x/x+.24)? I had that idea earlier but I wasn't sure if that would even be close.

    If that's right, i get a height of 2.536
     
  18. Oct 10, 2007 #17

    learningphysics

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    height = (d+0.24)sin32

    so now just use the energy equations...
     
  19. Oct 10, 2007 #18
    Okay, how about calculating velocity. Or would I instead use the potential energy for a spring.
     
  20. Oct 10, 2007 #19

    learningphysics

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    You don't need velocity. velocity at the bottom is just 0 ie no kinetic energy. just use mgh = (1/2)kx^2

    solve for d.
     
  21. Oct 10, 2007 #20
    So my equation should look like

    3*9.8*(sin(32)*(d+.24))=1/2*439*.24
     
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