Calculating Spring Constant for a Bungee Cord: Which Method is Correct?

In summary, the conversation discusses methods for calculating the spring constant of a bungee cord. In the first method, the equation k= Fx / (x-x0) is used, while the second method uses the equations PE=mgh, KE=0.5mv^2, and F=-kx. The second method gives a spring constant of 49, while the first method gives a spring constant of 19.6. The experts in the conversation determine that the second method is correct, and explain that the first method is incorrect because it does not take into account the jumper's weight and the spring's unstretched length.
  • #1
the_awesome
56
0
Hey guys. I'm trying to figure out the spring constant of a bungee cord.
Weight = 80kg
Gravity = 9.8m/s/s
Length of cord = 10m
Length of jump = 50m

I've been shown two different methods. And I don't know which one is correct because they give two different answers :/

1.
k = F/x = mg/x
Also knows as : k=Fx/x-x0

So i substitute: k= (80kg x 9.8) / (50-10)
k= 19.6 N/M

2.

PE=mgh
KE=0.5mv^2 = 0.5kx^2
F= -kx

mgh=0.5kx^2
80 x 9.8 x 50 = 0.5k x 40^2
therefore k= 49



So which method is correct? One gets 19.6, the other gets 49.
 
Physics news on Phys.org
  • #2
Is this a homework question? If the bungee cord stiffness is constant (which is generally not the case), then the second method is correct. The first method is incorrect.
 
  • #3
No it's not a hmwk question. I was at a bungee jumping place and I needed to do some calculations. The fact is that the 1st method is basically the same as the spring constant equation.

k= Fx / x - x0
However I do not know what x or x0 is so I'm confused.

So your saying that the 2nd method would hold true, if I was at a bungee jumping place, calculating the spring constant of the cord? If so then thankyou
 
  • #4
the_awesome said:
No it's not a hmwk question. I was at a bungee jumping place and I needed to do some calculations. The fact is that the 1st method is basically the same as the spring constant equation.

k= Fx / x - x0
However I do not know what x or x0 is so I'm confused.
F = k(x-x0) is just Hooke's law for springs. x-x0 is how much the spring stretches from its unstretched length.

When the jumper is at the lowest point, the upward force of the spring is given by that equation, but that force does not equal the weight of the jumper. (The spring force is greater than the jumper's weight at that point.) As nvn says, the first method is incorrect.
 
  • #5
Okay thankyou guys. 49 does seem like a massive spring constant. But thanx
 

What is a spring constant?

A spring constant, also known as a force constant or stiffness, is a measure of the stiffness of a spring. It is the amount of force required to stretch or compress a spring by a certain distance.

How is spring constant calculated?

The spring constant can be calculated by dividing the force applied to the spring by the change in length or displacement of the spring. It can also be calculated by dividing the elastic potential energy stored in the spring by the change in length.

What is Hooke's Law and how does it relate to spring constant?

Hooke's Law states that the force required to stretch or compress a spring is directly proportional to the displacement or change in length of the spring. The constant of proportionality is the spring constant. This means that the spring constant is a measure of how much force is needed to change the length of a spring by a certain amount.

What factors can affect the spring constant?

The spring constant can be affected by the material and shape of the spring, the number of coils, and the diameter of the wire used to make the spring. It can also be affected by the temperature and external forces applied to the spring.

How is spring constant used in real-world applications?

Spring constant is used in various real-world applications such as in the design of suspension systems for vehicles, shock absorbers, and springs used in mattresses and furniture. It is also used in scientific experiments to measure and understand the properties of materials and systems.

Similar threads

Replies
16
Views
4K
Replies
44
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
368
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Mechanics
Replies
4
Views
4K
Replies
26
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
Back
Top