Anonymous
Anonymous asked in Science & MathematicsPhysics · 5 months ago

Physics Question?

See picture for first part of problem and pictures.

II. You are to find:

a) The necessary un-stretched length of the bungee cord.

b) The maximum speed and acceleration achieved during the jump.

c) The spring constant of the bungee cord in relation to the jumper’s

 mass.

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1 Answer

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  • NCS
    Lv 7
    5 months ago
    Favorite Answer

    Welcome to Yahoo!Answers.

    All quantities are in SI units.

    person's GPE becomes spring PE:

    m*g*(25 - 2) = ½kx²

    where x is the maximum elongation of the cord,

    h - 2 = L + x

    where L is the unstretched length

    At equilibrium, the stretch z is due only to the person's weight:

    mg = kz

    and

    x = z + 8

    so

    mg = k(x - 8)

    k = mg / (x-8)

    m*g*23 = ½*[m*g/(x-8)]*x²

    m*g cancels

    23*2 = x² / (x-8)

    46(x - 8) = x²

    x² - 46x + 368 = 0

    is quadratic with roots at

    x = 10.3 m ← solution

    and x = 35.7 m ← outside of acceptable range

    a) so

    L = 23m - x = 12.7 m

    b) the maximum speed is achieved just when the spring force is equal to the weight:

    z = x - 8m = 2.3 m

    KE = initial GPE - final GPE - SPE

    KE = m*g*(12.7+2.3) - ½*[m*g/2.3]*(2.3)²

    KE = m*g*(15 - 1.15) = m*g*14

    and of course KE = ½mv²

    so after canceling the mass we have

    ½v² = g*14 = 9.8*14 = 137

    v² = 274

    v = 17 m/s

    c) k = mg / (x - 8) = m*9.8 / 2.3 = 4.26*m

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