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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: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant . Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward. 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. 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. solve d using the measurements that you were given. And; 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 ? Ignore all dissipative forces. (k in terms of L, h, m and g)

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Answer:

The spring constant of the bungee rope is given as

[tex]k = \frac{2mgh}{(h - L)^2}[/tex]

Explanation:

If Kate just touches the water surface while it is in downward motion

So the extension in the length of the rope is given as

[tex]x = (h - L)[/tex]

now we can use the energy conservation

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

so we have

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

so we have

[tex]k = \frac{2mgh}{(h - L)^2}[/tex]

The spring constant k in terms of L, h, m and g is;

k = 2mgh/(h - L)²

We are told that the bungee chord with length stretches as it behaves like an ideal spring.

Now, it stretches over a height h above water. Thus, the extension of the bungee chord will be;

x = h - L

We are told that Kate touches the surface of the water on her first downward trip. Thus, her kinetic energy is;

KE = ½kx²

Where;

k is spring constant.

x is extension of spring

While her potential energy at the start is;

PE = mgh

Now, from conservation of energy, we know that;

Potential energy = kinetic energy.

Thus;

mgh = ½kx²

Making k the subject, we have;

k = 2mgh/x²

From earlier, we saw that x = h - L

Thus;

k = 2mgh/(h - L)²

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