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Simple mechanical energy question [gr.11]

  1. Mar 22, 2004 #1
    [urgent] simple mechanical energy question [gr.11]

    Several children, pretending they are playing in the jungle, suspend a rope from an overhead tree limb. A child of mass 40 kg running at 8.0m/s grabs the rope and swings off the level ground.

    a) What maximum height does the child reach?
    b) How fast would a 30 kg child have to run to reach the same height as the 40 kg child?


    ½mv1²+mgh1 = ½mv2²+mgh2

    a) m=40kg, v1=8.0m/s, h1=0m, g=9.8m/s², v2=0m/s (at max height v=0)

    h2=?

    ½mv1²+mgh1 = ½mv2²+mgh2

    ½mv1²= mgh2

    h2 = 3.27m = 3m (is this correct?)

    im really confused with B... coz i got 8m/s... it doesn't make sense to me how they can run at the same speed and achieve the same height??!!

    this is really urgent... so please help me!! >_< thanks!
     
    Last edited: Mar 22, 2004
  2. jcsd
  3. Mar 23, 2004 #2
    A)
    I would say: the KE at the point of him grabbing it = the PE at his peak(all his KE has become PE) so KE=PE so ½mv²=mgh substituting:
    .5*40*8²=40*9.8*h
    1280=392h
    h=3.265m

    B)change 30 for 40 above and v is unknown
    .5*30*v²=30*9.8*3.265
    15v²=959.91
    v²=63.994
    v=7.999m/s

    yes, it is the SAME v for both! Why? because the m cancels out on both sides of the equation: the m in ½mv² and the m in mgh divide out!
     
    Last edited: Mar 23, 2004
  4. Mar 23, 2004 #3
    thank you!! ^^
     
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