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PE and KE, mass sliding down incline into a spring (no friction)

  1. Jul 6, 2015 #1
    1. The problem statement, all variables and given/known data
    In the figure, a block of mass m = 15 kg is released from rest on a frictionless incline of angle θ = 26°. Below the block is a spring that can be compressed 3.1 cm by a force of 380 N. The block momentarily stops when it compresses the spring by 4.0 cm. (a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches the spring?

    http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c08/fig08_41.gif

    2. Relevant equations
    20150706_180858_zpsnlsvs73a.jpg


    3. The attempt at a solution
    20150706_181021_zps1nvxryei.jpg
    20150706_181132_zps4gpctm2a.jpg
     
    Last edited: Jul 6, 2015
  2. jcsd
  3. Jul 6, 2015 #2

    Nathanael

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    Look at your expression for Ws at the bottom... you used (0.4m)^2 instead of (0.04m)^2
     
  4. Jul 6, 2015 #3
    Hey Nathanael, thanks. let me try this again
     
  5. Jul 6, 2015 #4
    when solving for part b im using K=1/2 mv2 , where K=9.8065, m=15 and i get v=1.143
     
  6. Jul 6, 2015 #5
    using kinematic eq. v2=2a(x-x0) where a=4.296 and x=0.15218. I got v=0.4461 m/s but not correct
     
  7. Jul 6, 2015 #6

    Nathanael

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    Kinematics equations were derived with the assumption that acceleration is constant, so it doesn't apply to this situation. (Their use is pretty limited.)

    I think this is the trap they expected you to fall into. Be careful with the energy: is there any other form of energy involved?
     
  8. Jul 6, 2015 #7
    I dont see why kinematic eq wont work, because its asking for velocity before it even touches the spring. so a should be constant?
    There could be a potential energy of gravity on the block. U=mgh
     
  9. Jul 6, 2015 #8

    Nathanael

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    You're right, I wasn't thinking about what you were doing o0) sorry
    Your mistake is that x you used is the total distance including the distance compressed. You just want the distance before it touches.

    Right. As the spring is compressed gravity is still doing work, so the kinetic energy of the block right before it touches is not equal to the total energy of the system.
     
  10. Jul 6, 2015 #9
    wouldnt the total distance be .15218+.04?

    Edit. nevermind its .15218-.04 for the dist before spring

    Edit2. after plugging in .11218 i still am doing something wrong when solving for v

    Edit3. ok i got it. i was trying to square .11218 in the kinematic equation, i dont know why i did that.
     
    Last edited: Jul 6, 2015
  11. Jul 6, 2015 #10

    Nathanael

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    Since you got it I will tell you the energy method of doing it: ##E_{kinetic}=E_{total}-0.04mg\sin(26°)## and you already determined Etotal for an earlier part.

    What you used in post #4 was Ekinetic=Etotal
     
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