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Conservation of energy, PE & KE problem. help?

  1. Oct 19, 2012 #1
    Conservation of energy, PE & KE problem. help??

    a block is sent sliding down a frictionless ramp. Its speeds at points A and B are 2.00 m/s and 2.60 m/s, respectively. Next, it is again sent sliding down the ramp, but this time its speed at point A is 4.00 m/s. What then is its speed at point B?


    I understand the first step is to calculate the KE

    KE at pt A:
    KEa = (1/2) m(2)^2
    KEa= 2m*
    calculating the KE at pt B:
    KEb = (1/2)mv^2
    KEb = 3.38m

    This part from the solution guide that I'm confused about:
    "the difference between the KE at pt B and at pt A is the gravitational potential energy, therefore
    Eg = KEb - KEa
    Eg = 3.38 - 2m
    Eg = 1.38m (m is the mass)"

    I don't understand this part!

    So PE = KEa - KEb??

    Why is that?


    And then (this part I also do not understand)
    KEa2 = 1/2m(4.00)^(2) = 8m

    To get the answer it would be

    8m+1.38m=9.38m

    1/2mv^(2) = 9.38m
    V= 4.33 m/s



    I am very confused. Can someone please explain to one to me. Thanks !
     
  2. jcsd
  3. Oct 19, 2012 #2

    Doc Al

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    Staff: Mentor

    Re: Conservation of energy, PE & KE problem. help??

    Point A must be higher than point B. As the block lowers to point B, some PE transforms to added KE.

    Mechanical energy is conserved:

    KEA + PEA = KEB + PEB

    Rearranging, that becomes:
    KEB - KEA = PEA - PEB
     
  4. Oct 19, 2012 #3

    TSny

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    Homework Helper
    Gold Member

    Re: Conservation of energy, PE & KE problem. help??

    Conceptually (no numbers for now), how is the change in KE of the block related to the change in PE when sliding from a to b? [EDIT: Doc Al posted while I was still typing. Beat me to the punch :)]
     
  5. Oct 19, 2012 #4
    Re: Conservation of energy, PE & KE problem. help??

    While I was reading your post, it didn't occur to me to use the method you mentioned. If you are keen to take the energy approach and it works for you, by all means do!

    However, the answer you quoted can be reached by just using Newton's laws of motion. Perhaps it would be easier for you to see what is going on in the problem this way?

    Although it seems at first that you don't have enough information to fill in all the blanks (u,v,a,s and t) consider that you might be able to include some of these generally (just as letters), because they can be cancelled out later.
     
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