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Work energy and power.

  1. May 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Ima Scaarred (m=56.2 kg) is traveling at a speed of 12.8 m/s at the top of a 19.5-m high roller coaster loop.

    a. Determine Ima's kinetic energy at the top of the loop.
    b. Determine Ima's potential energy at the top of the loop.
    c. Assuming negligible losses of energy due to friction and air resistance, determine Ima's total mechanical energy at the bottom of the loop (h=0 m).
    d. Determine Ima's speed at the bottom of the loop.



    2. Relevant equations
    PE=MGH
    KE=0.5*MV^2



    3. The attempt at a solution
    i tried to find b

    pe=0.5*56.2*12.8

    =14,027.52


    help me understanding the question.

    the answers are.

    • Show Answer
    a. 4.60 x 103 J
    b. 1.07 x 104 J
    c. 1.53 x 104 J
    d. 23.4 m/s
     
  2. jcsd
  3. May 27, 2016 #2

    billy_joule

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    Science Advisor

    Looks like you've mixed your two relevant equations together.

    your relevant equation says pe = mgh, but that's not what you've done, what is m? g? h?
    First identify all your variables so you know where they'll go in your equations.

    Also, your arithmetic is wrong, 0.5*56.2*12.8 does not equal 14,027.52
    Show your units with your working, and be specific on exactly where you're stuck/ what you don't understand, it makes it easier to help.
     
  4. May 27, 2016 #3
    PE= mass*gravity*height

    i want to imagine the picture of that question so that it would be easy for me , i want to understand the question
     
  5. May 27, 2016 #4
    sorry the answer of pe=0.5*56.2*12.8
    is 359.68
    but it is also wrong....

    is there any role of friction?
    or use of this formula
    Pi+Ki=Pf+Pi
     
  6. May 27, 2016 #5
    All the ANswers i have got!! thank you bilijoule i was making big mistake,,,,
     
  7. May 27, 2016 #6

    billy_joule

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    Great, good work.
     
  8. May 28, 2016 #7
    thankyou\
    but i m still trying to find the speed at the bottom of the loop .

    KE=KE
    4600=0.5*56.2*v^2
    4600=28.1*v^2
    4600/28.1=v^2
    v^2=163.7
    square root both sides
    v=12.8

    so this way i got the initial velocity but how to get the speed at the bottom?
     
  9. May 28, 2016 #8

    billy_joule

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    When Ima gets to the bottom, the potential energy she had at the top has been converted to kinetic energy via Conservation of Energy.
    This should be somewhat intuitive, ever ridden your bike down a hill? You don't even need to pedal to gain kinetic energy..it's coming from somewhere else..
    The higher the hill the faster you'll be going at the bottom...
    The relevant equation:
    EKi + EPi = EKf + EPf

    (which is what I think your earlier equation in post #4 was supposed to be?
    )

    Solving for EKf is the answer to c), then solving EKf = 1/2 mv2 for v will give the answer to d)
     
  10. May 28, 2016 #9
    v=23 m/s !! thanks for the equation .
    KEf=1/2mv^
     
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