Restoring Forces and Hookes law

In summary, Kate, a bungee jumper, wants to jump off a bridge that spans a river, with a mass m at a height h above the water. The bungee cord, with length L, behaves as an ideal spring with spring constant k. When Kate falls, the bungee cord first straightens and then stretches. The spring constant can be expressed as k=(mg)/(h-L). If Kate just touches the surface of the river on her first downward trip, the spring constant can be expressed as k=1/2(mg)/(h-L)^2.
  • #1
Emendez3
5
0
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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  • #2
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?
 
  • #3
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.
 
  • #4
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
 
  • #5
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.
 

FAQ: Restoring Forces and Hookes law

1. What are restoring forces?

Restoring forces are forces that act on an object to bring it back to its original position after it has been displaced. These forces are present in many natural phenomena, such as the movement of a spring or the oscillation of a pendulum.

2. What is Hooke's law?

Hooke's law is a principle in physics that states the force exerted by a spring is directly proportional to the amount it is stretched or compressed. This means that the force and displacement are directly related and can be represented by a linear equation: F = -kx, where F is the force, k is the spring constant, and x is the displacement.

3. How does Hooke's law relate to restoring forces?

Hooke's law is directly related to restoring forces because it explains the relationship between the force applied to an object and the resulting displacement. In other words, Hooke's law describes the magnitude of the restoring force that is present when an object is displaced from its equilibrium position.

4. What factors affect the strength of restoring forces?

The strength of restoring forces is affected by several factors, including the stiffness of the material, the amount of displacement, and the distance between the object's equilibrium position and its displaced position. Additionally, the type of force (e.g. tension, compression, etc.) and the shape of the object can also impact the strength of restoring forces.

5. How is Hooke's law used in real-world applications?

Hooke's law has many practical applications, including in the design of springs and shock absorbers. It is also used in materials testing to determine the stiffness and elasticity of materials. Additionally, Hooke's law is essential in understanding the behavior of many natural phenomena, such as the movement of tectonic plates and the behavior of elastic materials.

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