Restoring Forces and Hookes law

AI Thread Summary
Kate, a bungee jumper, steps off a bridge, and the discussion revolves around calculating the spring constant k of the bungee cord. The forces acting on Kate are analyzed, noting that at the lowest point of her fall, the forces are not balanced, leading to zero velocity, similar to a thrown ball at the peak of its trajectory. The potential energy at the start of the jump is converted into the work done by the bungee cord as it stretches. The correct relationship derived is m*g*h = 1/2*k*(h-L)², which allows for the calculation of k in terms of L, h, m, and g. This analysis emphasizes the importance of understanding energy conservation and force balance in bungee jumping scenarios.
Emendez3
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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 k.
* 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.


Using:

Fsp=Fg=mg
and Fsp=kdelta(s)
where
Fsp= force of spring
m=mass
k=spring constant

i arrived at
k=(mg)/(h-L)

but its wrong it says:
At this lowest point, forces are not balanced. If they were, Kate's momentum would carry her farther down, into the water. She actually has zero velocity at this point, much like a thrown ball does at the top of its trajectory.
But i don't know what that means
 
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Emendez3 said:
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 k.
* 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.


Using:

Fsp=Fg=mg
and Fsp=kdelta(s)
where
Fsp= force of spring
m=mass
k=spring constant

i arrived at
k=(mg)/(h-L)

but its wrong it says:
At this lowest point, forces are not balanced. If they were, Kate's momentum would carry her farther down, into the water. She actually has zero velocity at this point, much like a thrown ball does at the top of its trajectory.
But i don't know what that means

What is the question? Find k in terms of H,L,m,g?
 
LowlyPion said:
What is the question? Find k in terms of H,L,m,g?

oh, 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.
 
Emendez3 said:
oh, 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.

Well think about it then.

When she jumps she has m*g*h in PE. And when she falls the bungee doesn't begin to retard her until she has fallen L.

So that means that the work done by the bungee to just stop her at h below the top must be

m*g*h = 1/2*k*(h-L)2
 
LowlyPion said:
Well think about it then.

When she jumps she has m*g*h in PE. And when she falls the bungee doesn't begin to retard her until she has fallen L.

So that means that the work done by the bungee to just stop her at h below the top must be

m*g*h = 1/2*k*(h-L)2

Oh ok Thank you very much.
 
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