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Conservation of energy?

  1. Apr 30, 2007 #1
    1. The problem statement, all variables and given/known data
    A 200g block is pushed against a horizontal spring of constant 200 N/m until it is compressed 15cm. When the mass loses contact with the spring it moves over a horizontal rough patch of surface of length 50cm and coefficient of friction 0.2. After passing this patch the block continues to slide up a frictionless ramp of angle 30degrees.

    A)How far does the black rise on the ramp after the first pass?

    2. Relevant equations
    What i tried to do was solve for the PE of the spring and then the energy lost from the patch of rough ground then solve for the resulting gravational PE.

    so PE_spring - E_friction = PE_gravity

    3. The attempt at a solution
    Really didn't get any good attempted solutions but would just like to know if Im on the right path or if Im completely wrong. -Thanks Tom
  2. jcsd
  3. Apr 30, 2007 #2
    Looks good to me. Remember to keep all of your units compatible. (ie, don't mix 9.8 m/s^2 with 15cm)
  4. Apr 30, 2007 #3
    quick question when it says "horizontal spring of constant 200 N/m until it is compressed 15cm" does the 200N/m = k or do i plug in 200N/m = -k(15cm) to get K

    -Thanks TOm

    This is what i did so far

    U(x)=-100x^2 = 0 - (-100*(-.15)^2 = 2.25


    F=.2(9.8)(200) = 392
    E=392(.5)= 196

    I think I'm not solving for energy of friction correctly because if this is right then it isn't even going to pass the patch of friction
    Last edited: Apr 30, 2007
  5. Apr 30, 2007 #4
    quick answer:
    The first.
    k=200N/m, x=0.15m.
  6. Apr 30, 2007 #5
    really lost now because i can't get past this 2.25 - 196 i know one of these numbers is way off. do i multiply 2.25 by the mass?
  7. Apr 30, 2007 #6
    How did you obtain those two numbers? What are they meant to represent?
    edit: just saw your edit. I rescind the question.
    Last edited: Apr 30, 2007
  8. Apr 30, 2007 #7

    Good evaluation, there.
    You're given the mass of the block in grams.
  9. Apr 30, 2007 #8
    o thats a mistake so if i use .2kg i get
    E_friction= (.5)(.2)(.2)(9.8) = .196

    2.25-.196 = mgy Y=2.054/mg = 1.04796
    y= 1.04796
    (so that should be the anwser if i did everything right)
  10. Apr 30, 2007 #9
    That's what I got!

    And if I had to guess, that's what I'd enter as the answer (with correct number of significant figures, and including the units). But there is a slight chance that the question wants the distance travelled along the surface of the ramp... but from the phrasing of the question, it sounds like they want the vertical height, so you're done.
  11. Apr 30, 2007 #10
    Thanks for all the help now i just got few quick questions for the parts b,c,d (I think i got the answers already)

    B) The mass comes down and goes over the rough patch again on the way back to the spring. how many passes will the mass make over the patch before running out of energy?

    My answer:
    Initial energy = 2.25
    Energy lost per pass= .196
    so 2.25-.196x=0
    x=number of passes = 11.4796

    C)where along the rough patch will the block finally stop?
    My answer:
    it will end .4796 through the last trip so
    (.4796)(.5)=final distance =.2398

    D)how long would the patch have to be so that the block never reaches the ramp?
    My answer:
    if it makes 11.4796 passes then the patch must be 11.4796 times its current lenght to just make one pass
    (11.4796)(.5)= 5.7398

    Well those are the answers i got for the the last few parts, not really sure if there right though, they just kinda seem to make sense to me.
  12. Apr 30, 2007 #11
    o and when i check the friction using the distance 5.7398 the force comes out to equal 2.25 which makes me think it is right since i want the energy to = 0
  13. Apr 30, 2007 #12
    I'm not sure about your answer to part C.
    Thing to consider (which I haven't considered, by the way, but I don't think that you have, either):

    Does the block end coming from the ramp or coming from the spring?
    In other words, 0.2398meters from where?
  14. Apr 30, 2007 #13
    I was thinking about that and thought i had it right but i just relized i think it is coming back from the ramp.

    0 -> 1 = from spring to ramp
    1 -> 2 = from ramp to spring
    2 -> 3 = from spring to ramp

    so going to an odd number = going towards ramp
    and going to even number = going toward spring

    meaning i did mine wrong and it should be .5-.2398= .2602m over the rough patch coming from the spring side

    Thanks for pointing that out, i had already thought about it but thought i did it right the first time. Thanks for all your help!
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