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Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 decade ago

Physics problem help please? Energy problem!?

Im stuck on this one physics problem, need all work to it so i can get ready for my test. Thanks so much :] 10 points

Chimpo the chimp, with a mass of 80 kg, runs and grabs a tire swing with a mass of 20 kg and both swing up to height of 2 m. How fast was chimpo running when he grabbed the tire?

4 Answers

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  • ?
    Lv 4
    1 decade ago
    Favorite Answer

    Its quite simple

    Chimpo has Kinetic energy when he reaches the tire, and chimpo and the tire have P)otential energy when they have risen to the height of 2metres

    Energy cannot be destroyed only converted

    So total KE of chimpo = total PE of chimpo and the tire

    0.5 x 80 v^2 = (80 + 20) x 10 x 2

    40 x v^2 = 2000

    V^2 = 2000/40 = 50

    v = sqrt 50 = 7.07 ms is Chimpo's velocity

  • 1 decade ago

    Using conservation of energy, you get that Ek=Ep (Kinetic energy= potential energy)

    1/2mv^2 = mgh

    1/2 x 80 x v^2 = (80+20) x (-9.8) x 2

    v^2 = 100 x (-9.8) x 2 x 2/80

    v^2 = 49

    v = √49 = 7 m/s

    That's your answer!

    By the way the value of g is 9.8 not 10!

  • Anonymous
    5 years ago

    To all the students out there with teachers to impress, Physics is easy to do, just follow these steps: 1: You get a poisonous box 2: You put a cat in that box 3: You let it stay in the box And that's the way you do it!

  • 1 decade ago

    Let

    Po = initial momentum of Chimpo

    P = final momentum of (Chimpo + tire)

    Then

    (1) ..... Po = Mc*Vo

    where Mc = mass of Chimpo, Vo = initial speed of Chimpo

    (2) ..... P = ( Mc + Mt ) * Vct

    where Mt = mass of tire, Vct = initial speed of Chimpo and tire

    By conservation of linear momentum (Po = P)

    (3) ..... Mc*Vo = ( Mc + Mt ) * Vct

    Now

    (4) ..... Eo = (0.5)*( Mc + Mt ) * Vct^2 = initial energy of Chimpo

    and tire at bottom of swing

    (5) ..... E = ( Mc + Mt )*g*H = final energy of Chimpo and tire at top of swing

    where H = 2.0 m

    By conservation of energy (Eo = E)

    .......... (0.5)*( Mc + Mt ) * Vct^2 = ( Mc + Mt )*g*H

    (6) ..... Vct^2 = 2*g*H

    From (3)

    (7) ..... Vct = Mc*Vo / ( Mc + Mt )

    Substituting (7) in (6)

    .......... ( Mc^2)*(Vo^2) / [ ( Mc + Mt )^2 ] = 2*g*H

    .......... Vo^2 = 2 * g * H * [ ( Mc + Mt )^2 ] / ( Mc^2)

    (8) ..... Vo = SQRT { 2 * g * H * [ ( Mc + Mt )^2 ] / ( Mc^2) }

    Substituting the values

    .......... 2 * g * H = 2 * (9.8) * (2.0) = 39.2

    .......... [ ( Mc + Mt )^2 ] / ( Mc^2) = [ ( 80 + 20 )^2 ] / [ ( 80 )^2) ] = 1.5625

    .......... Vo = SQRT { ( 39.2 ) * ( 1.5625 ) } = 7.826

    .......... 7.8 m/s = speed of Chimpo when he grabbed the tire

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