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Potential Energy and Conservation of Energy Problem

  1. Mar 1, 2009 #1

    1. The problem statement, all variables and given/known data
    A pendulum consists of a 4.0 kg stone swinging on a 3.2 m string of negligible mass. The stone has a speed of 7.7 m/s when it passes its lowest point.

    2. Relevant equations
    (a) What is the speed when the string is at 62° to the vertical?
    (b) What is the greatest angle with the vertical that the string will reach during the stone's motion?
    (c) If the potential energy of the pendulum-Earth system is taken to be zero at the stone's lowest point, what is the total mechanical energy of the system?

    3. The attempt at a solution
    I'm all confused >.< I figured there wold be max KE in the middle. Then there would be some KE & PE at the 62 degree part.. but idn what to do.. ugh I tried Ei = Ef so then Ei = .5mv2 but what would Ef be... idn.. i was thinking about centripetal force... idn i'm hopeless >.<
     
  2. jcsd
  3. Mar 1, 2009 #2
    ME [Mechanical Energy] = KE [Kinetic Energy] + PE [Potential Energy]
    At the stone's lowest point you could take the PE to be 0J (since it could be taken as the reference level); the KE at that point would then be the total mechanical energy of the system. At 62deg to the vertical, find the height above the reference level to solve for the PE. The difference between the ME and PE at that point is the KE. Then solve for the speed. The greatest angle would occur at the height where KE = 0.
     
  4. Mar 1, 2009 #3

    hmm...
    okie...
    I was thinking of finding the height change, but I confuse myself sometimes. I couldn't figure out how to get it. So, you're supposed to do 3.2cos(62) = 1.502
    And then do 3.2 - 1.502 = 1.698 which will equal the height change...
    So now..
    E (which equals the KE initial) = KE + PE
    so plug in numbers 118.58 = (.5)(4)v^2 + (4)(9.8)(1.698)
    v = 5.10 m/s

    ooh yay!! tehehe gots it!

    and then basically work backwards for the angle
    so E = mgh

    and then for C.. it's just the E we've been working with

    thankyou very much! ^_^ highly appreciated XD

     
    Last edited: Mar 1, 2009
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