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A block on a frictionless circular ramp

  1. Apr 8, 2014 #1
    1. The problem statement, ramp. variables and given/known data
    A block with mass m=5kg is placed at position A and given an initial velocity Va=2m/s Down a frictionless circular ramp. Between positions B and C it travels over a flat rough surface having a coefficient of kinetic friction Mk=.25. Finally it travels over a flat frictionless surface and contacts am uncompressed so spring with spring constant k=1250N/m. The block comresses the spring at distance Delta (X) and comes to rest at position D.


    2. Relevant equations

    A) beginning at position A use conservation of energy to calculate the blocks velocity when it reaches position B

    B) begimning at position B use conservation of energy to calculate the block's velocity when it reaches position C . Don't forget friction!!!!

    C) beginning at position C use conservation of energy to calculate the distance Delta(X) that the spring is compressed when the block reaches position D

    D) starting over at position A use conservation of energy directly between positions A and D and show that you get the same distance Delta(X) as part c) without having to calculate the velocities at position B and C

    3.Please I'm super lost and have tried everything can someone solve and show me?!!!
     
    Last edited: Apr 8, 2014
  2. jcsd
  3. Apr 8, 2014 #2

    BiGyElLoWhAt

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    Gold Member

    how tall (or long with angle) is the ramp?
    CoE:
    A)[itex]\frac{1}{2} mv_{initial}^2 +mg\Delta h = \frac{1}{2} mv_{final}^2[/itex] where h is either given or = [itex]dsin(\theta)[/itex]
    B) do the same thing, but include friction as part of the work term [itex]W = mg\Delta h - \mu N\Delta x[/itex]
    C) again, do the same thing, I'm assuming flat means level? so
    [itex]\frac{1}{2}mv_{InitialAfterC}^2 = \frac{1}{2}k\Delta x^2[/itex] W=0, no friction, and mg sin theta = 0, so all kinetic energy goes into compressing the spring (potential)
    i'm not crunching the numbers for you, but that should work, give it a shot.
     
    Last edited: Apr 8, 2014
  4. Apr 8, 2014 #3
    Ramp is 3m!!!
     
  5. Apr 8, 2014 #4
    Can you please since it I have no ideas how to change the formula to get velocity final
     
  6. Apr 8, 2014 #5

    BiGyElLoWhAt

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    Gold Member

    what do you mean? i gave you the equation. plug in 1/2, plug in m, plug in initial velocity. that gives you the first term, which is a number. then plug in m, plug in g, plug in h. that gives you the second term which is a number. then plug in 1/2, plug in m, and keep v final ^2 which gives you some number times v final^2.

    you get some number = some number times v final^2 . plug it into your calulator, its a 16 =4x^2 type equation, and i'm confident you can solve it. once you get v final for A), use that for v inital for B) and so on.

    this forum is for homework help not for us to give you the answers, which i pretty much did already.
     
  7. Apr 8, 2014 #6
    I tried I got stuck help me :(
     
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