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Energy Question. I REALLY NEED HELP

  1. Nov 7, 2007 #1
    [SOLVED] Energy Question. I REALLY NEED HELP

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

    A block of 1 kg is sent sliding down a frictionless curved ramp. The block reaches the horizontal and hits a rough surface where a frictionless force of 2N acts for 5 meters. The block continues on, going up the frictionless incline. The height of the ramp is also 5 meters.

    a) Wwhat is the energy at point A? (points A is at the beginning of the ramp)

    b) What is the velocity of the block at point B? (Point B is at the bottom of the ramp but not in the center, it is right before the disturbance--or 2N of force--starts to act)

    c) What is the velocity at point C? (Point C is after the disturbance of 2N)

    d) What is the height up the incline reached by the box?

    2. Relevant equations

    a) Ug = mgh

    b) Ug=mgh and K= m(vsquared) / 2
    so you set them equal to each other.

    c) W=Fs FORCE AND DISPLACEMENT
    i DONT KNOW HOW TO DO THIS ONE.

    D) I REALLY DON'T KNOW HOW TO DO THIS PART?


    3. The attempt at a solution

    a) 50 J
    B) 10 m/s
    C) ??
    D) ???
     
    Last edited: Nov 7, 2007
  2. jcsd
  3. Nov 7, 2007 #2
    Please Help I Really Need Help On This, Anyone Can Help
     
  4. Nov 7, 2007 #3

    PhanthomJay

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    For c, the work done by the friction force will change (reduce) the KE of the block. You can use the work-energy theorem. For d, then proceed as you did in part b.
     
  5. Nov 7, 2007 #4
    but i dont get how you can reduce the KE
    like what equation do you use for C?????????

    like do you do 50 J - 10J= 40J and then set 40J = m (v squared) / 2????
     
  6. Nov 7, 2007 #5

    PhanthomJay

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    Yes, but get familiar with the work energy theorem:
    [tex]W_T = \Delta KE[/tex] where [tex]W_T[/tex] is the work done on the object by all forces acting on it. Note that it has a negative value in this case, so that your equation reads
    [tex] -10 = KE_f - 50[/tex] thus
    [tex]KE_f = 40J[/tex]
     
  7. Nov 7, 2007 #6

    Astronuc

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    Well I'm puzzled by the problem statement.

    A curved ramp and an incline (which is normally a straight ramp)?

    Does the block start at 5 m elevation on the curved ramp? In that case, mgh = 49 J assuming g = 9.8 m/s2, and then the KE = mgh at the horizontal point, i.e. gravitational potential energy is transformed into kinetic energy.

    The 2 N over 5 m applies 10 J of work against the block. Is the 5 m over which the 2N is applied horizontal? If so, then the KE = KE (intial) - Work, as PhantomJay explained.
     
    Last edited: Nov 7, 2007
  8. Nov 7, 2007 #7
    OOOO okay thank you, see my physics teacher has not taught us that equation yet, but he gave us hw on it.
    anyways so now the V= 8.94 m/s. right ???? can you just check that??

    and for d) i do not get what i have to do still. do i do mgh = m (v squared) / 2. so it would be (1)(9.81)(h) = (1) (8.94--velocity from c) squared / 2
     
  9. Nov 7, 2007 #8
    it is not a straight ramp.

    yes the block starts at 5m elevation. the ramp is in like a U shape. the l part of the U is 5 meters and the curved part has point b and then the 5m of 2n force friction and then point C and then it has the rest of the l part of U.
     
  10. Nov 7, 2007 #9

    PhanthomJay

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    Yes that looks right, as long as the problem is as Astronuc stated and as I (and you?) have understood it. If you're not familiar with the work-energy theorem, you could have used Newton 2 to calculate the block's deceleration on the rough suface, then use the kinematic equations, but it takes a bit longer to do it this way. Also, make sure you are thoroughly familiar with the conservation of energy equation tha you used to calculate block speed at the bottom of the ramp or top of the incline.
     
  11. Nov 7, 2007 #10
    Okay, thank you so much for helping me PhanthomJay and Astronuc.
    This really helps. Hopefully, I will get a perfect score on this problem.
     
  12. Nov 7, 2007 #11

    PhanthomJay

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    I just realized that you used g=10m/s^2 in part b, to arrive at 50J , but g=9.81m/s^2 in part d. Be consistent no matter which value you use.
     
    Last edited: Nov 7, 2007
  13. Nov 7, 2007 #12

    Astronuc

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    It doesn't matter about the geometry after the 2N is applied. It does matter about the starting height initially, and whether the initial KE = 0, i.e. the block starts at rest.

    Basically the work done by the 2N force reduces the energy (kinetic energy of the block) by 10 J ( 2 N over 5 m). Then the gravitational potential energy on the incline must be 10 J less than the GPE from the start on the curved incline.
     
  14. Nov 7, 2007 #13
    so i did the calculations and for d) i got 4 meters. is that right? can one of you check that answer?

    THANKS SO MUCH,
    i appreciate it!
     
  15. Nov 7, 2007 #14
    and i am keeping the gravity at 10m/s i never changed that phanthomjay.
     
  16. Nov 7, 2007 #15

    PhanthomJay

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    Yep, using g=10m/s/s, 4 meters is the vertical height of the incline when the block stops, nice work.
     
  17. Nov 7, 2007 #16
    Thank You.
    I Could Not Have Done This Problem Without You!!!!!!!
     
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