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Energy/Work Problem

  1. Mar 20, 2005 #1
    I am stumped by a problem I was given to find the height of a thrown ball from the point of release. The problem is:

    During a contest that involved throwing a 7.0 kg bowling ball straight up in the air, one contestant exerted a force of 810 N on the ball. If the force was exerted through a distance of 2.0 M, how high did the ball go from point of release?

    I've been working for the past 15 minutes trying to set up an equation/s, but I can't figure out which way to take the problem. Any advice on how to arrange the given quantities would be appreciated.
     
  2. jcsd
  3. Mar 20, 2005 #2
    Hmm...I'm not exactly an expert...lol. But I'll give you my advice anyway.

    Draw a free body diagram on the ball. Find the sum of the forces (810 N going up, 700 N down)...remember W=F(distance)..
     
  4. Mar 20, 2005 #3
    Now, I'm pretty sure work=kinetic engergy. try to find the initial veloc. of the ball. then use kinematics to find max height.
     
  5. Mar 20, 2005 #4
    let me know if you understand what i'm saying...please show your work.
     
  6. Mar 20, 2005 #5
    I set it so that Work= Change in Kinetic Energy and got an answer I don't think is feasible. Fd=m(v squared)/2 (810)(2.0)=(7.0)(v squared)/2 v=square root(3240/7) which equals a velocity of 22 m/s after I plugged that into the Ke=Ug equation I got a height of 25 m.

    I have a feelign I'm making this problem a lot harder than it truly is.
     
  7. Mar 20, 2005 #6
    remember the sum of the force is just 810-700 (weight of ball)...=110
     
  8. Mar 20, 2005 #7
    At point of release KE = 1620J, PE = 0
    At point of impact KE = 0, PE = 1620J
    KE = .5mv^2
    PE = mgh
    PEi + KEi = PEf+KEf
    mgh+.5mv^2 = mgh+.5mv^2. Notice both sides equal 1620.

    mgh = 0 ______ .5mv^2 = 0

    .5mv^2 = mgh
    F*d = mv^2​
    .5v^2 = gh
    v^2 = F*d/m​

    F*d/(2m) = gh
    h = F*d/(2*m*g)
    h = (810)*(2)/(2*7*9.8)
    h = 1620/137.2
    h = 11.8m

    I hope this helps.
     
    Last edited: Mar 20, 2005
  9. Mar 28, 2005 #8
    Please correct me if I'm showing my stupidity here, but if the above statement is true, then would whozum's equation be incorrect in that
    "F*D=mv^2" or would it be F*D=.5mv^2,
    which would make it:

    F*D/m=gh --> h=F*D/m*g

    Thanks
    Chris
     
  10. Mar 28, 2005 #9
    For some reason when I wrote that, I had a reason that the 1/2 shouldnt be there, but now that I'm looking back at it, I have no idea why I omitted it.

    You are correct.

    [tex] KE = Fd = 1/2 mv^2 [/tex]
     
  11. Mar 28, 2005 #10
    thanks for clarifying that whozum
     
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