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Homework Help: Incline and Springs

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data
    A 50 g ice cube can slide without friction up and down a 30 degree slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 10 cm. The spring constant is 25 N/m. When the ice cube is released, what total distance will it travel up the slope before reversing direction?

    b)The ice cube is replaced by a 50 g plastic cube whose coefficient of kinetic friction is 0.20. How far will the plastic cube travel up the slope?

    2. Relevant equations

    3. The attempt at a solution
    I'm not a fan of incline problems and I honestly have no clue on how to start this question. Can someone please guide me through it and help me come up with a set of steps that will help me through future incline problems?

    1. The problem statement, all variables and given/known data

    The spring shown in the figure is compressed 50 cm and used to launch a 100 kg physics student. The track is frictionless until it starts up the incline. The student's coefficient of kinetic friction on the 30 degree incline is 0.15.

    a)What is the student's speed just after losing contact with the spring?
    b)How far up the incline does the student go?

    http://img530.imageshack.us/img530/6145/knightfigure1154nl2.th.jpg [Broken]

    2. Relevant equations

    3. The attempt at a solution
    I got part a) which was 14.1m/s using the law of conservation of energy, but as I mentioned before, I always get stuck on incline questions.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 28, 2007 #2
    um, ok well i got the answer to part a of question 1 which was 51cm, but im not sure how to include friction in the incline problem. can anyone please help?
  4. Oct 28, 2007 #3
    [tex]\frac {1}{2}m[v^{2} - v_{0}^{2}] + mg[y - y_{0}] + \frac {1}{2}k[x^{2} - x_{0}^{2}] = W_{F} + W_{ext}[/tex]

    so now there is no spring involved and at the top, we stop moving further and eventually slide down. so we take that instant to be v=0, our equation simplifies to

    [tex]\frac {1}{2}m[v^{2} - v_{0}^{2}] + mg[y - y_{0}] + = W_{F}[/tex]

    for part b
    Last edited: Oct 28, 2007
  5. Oct 28, 2007 #4
    i think i got confused

    is there a picture by chance? what is y_initial?
    Last edited: Oct 28, 2007
  6. Oct 28, 2007 #5
    in order to include friction, you must draw another force body diagram and solve for the Normal force

    you're already given kinetic friction, so just find normal and you're good to go!


    [tex]\phi=\mbox{angle between}[/tex]
    Last edited: Oct 28, 2007
  7. Oct 28, 2007 #6
    but with that do i add it to the potential energy, or only use the frictional force energy?
  8. Oct 28, 2007 #7
    sorry, but there is only a picture for the second one. Part b of both questions are similar so the method would be the same right?
  9. Oct 28, 2007 #8

    and the equation is still the same, just modified ... and i'm not sure what y-final or y-initial is, but of course y-initial should be taken as 0 and y-final, idk; and i'm sure you're solving for s.

    [tex]\frac {1}{2}m[v^{2} - v_{0}^{2}] + mg[y - y_{0}] + = \mu_{k}Ns\cos{\phi}[/tex]

    but in order to get N, you need a FBD ... which will most likely equal the weight.
    Last edited: Oct 28, 2007
  10. Oct 28, 2007 #9
    HOw did you get 51 cm for the first part?
  11. Oct 28, 2007 #10
    I didnt use a velocity, since there is only gravitational energy present in the equation. That's how i got 51. I know its right, my answer matches the correct answer
  12. Oct 28, 2007 #11
    I don't doubt that you are right. I just cant' seem to come to that value with my calculations.
  13. Oct 28, 2007 #12
    Er.. im not sure i keep getting 78 as my answer for how much he goes up the incline.. my h was 0 since it was he was at 0 before he goes up the incline, and my v was 14.1m/s and i solved for s but i got it wrong
  14. Oct 28, 2007 #13
    I got part B to that question..
  15. Oct 28, 2007 #14
    a = -g(ukcos0 +sino)
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