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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 k.
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.

Respuesta :

Answer:

[tex]y = L + \frac{mg}{k}[/tex]

Explanation:

After Kate jump and come to stop in equilibrium then we can say that in that position the net force on it will become ZERO

So we will have

[tex]F_{net} = 0[/tex]

[tex]mg - kx = 0[/tex]

[tex]x = \frac{mg}{k}[/tex]

So the position of Kate from the position of the bridge is given as

[tex]y = L + x[/tex]

[tex]y = L + \frac{mg}{k}[/tex]

onyebuchinnaji

The distance of Kate below is bridge for the stretched cord is [tex]d = L + \sqrt{\frac{2mgh}{k} }[/tex]

The given parameters;

  • mass of Kate, = m
  • height of the bridge, = h
  • length of the cord, = L

Apply the principle of conservation of energy and determine the extension of the cord when Kate steps off;

[tex]\frac{1}{2} kx^2 = mgh \\\\kx^2 = 2mgh\\\\x^2 = \frac{2mgh}{k} \\\\x = \sqrt{ \frac{2mgh}{k}}[/tex]

The distance of Kate below is bridge for the stretched cord is calculated as;

[tex]d = L + x\\\\d = L + \sqrt{\frac{2mgh}{k} }[/tex]

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