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Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls.
Assume the following:
A) The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant .
B) Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward.
C) Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle.
Use g for the magnitude of the acceleration due to gravity.
1) How far below the bridge will Kate eventually be hanging, once she stops oscillating and comes finally to rest? Assume that she doesn't touch the water.
Express the distance in terms of quantities given in the problem introduction.
Update:
I already found the answer to 1) to be:
d = L + (mg / k)
2) If Kate just touches the surface of the river on her first downward trip (i.e., before the first bounce), what is the spring constant k? Ignore all dissipative forces.
Express k in terms of L, h, m, and g.

Respuesta :

Answer:

1)   d = h- L - mg / k , 2)        k = 2 mg/h    (-1 +2 L / h)

Explanation:

1) Let's use the translational equilibrium equation

       [tex]F_{e}[/tex] -W = 0

       F_{e} = W

       k x = mg

       x = mg / k

         

from the statement of the exercise the height of the bridge, for the reference system in the river, is

        h = L + x + d

        x = h - L - d

we substitute

        h - L -d = mg / k

        d = h- L - mg / k

2) They ask us for the spring constant. For this part we can use energy conservation

Starting point. At the point before jumping

        Em₀ = U = m g h

Final Point. When it's hanging

        Em_f =   [tex]K_{e}[/tex] + U = ½ k x² + mg d

        Em₀ = Em_f

         mg h = 1 / k x² + m g d

in the exercise they indicate that Kate touches the surface of the river, so the distance d = 0

         mg h = ½ k x²

         k = 2 mg h / x²

       

we substitute the value of x

        k = 2mg  h / (h -L)²

        k = 2mg  h / (- L + h)²

we simplify the expression

       k = 2mg  h / [h² (1- L / h)²]

       k = 2m /h      (1- L / h)⁻²

In these jumps the bridge height is always greater than the length of the rope L / h <1, so we can expand the last expression

          (1- L / h)⁻² = 1 - 2 (1 -L / h) + 2 3/2!   (1 -L / h)² + ...

for simplicity let's keep up to the linear term, we substitute in the solution

       k = 2 mg/h    [1 - 2 (1- L / h)]

       k = 2 mg/h    (-1 +2 L / h)