Is my solution for the Bungee Tension Problem correct?

In summary, the conversation discusses finding the value of tension in a system using the equations for sum of forces and conservation of energy. The value of tension is found to be equal to 2mg, and the displacement of the cord is also determined to be 2mg/k. The conversation also considers the mass per unit length of the cord and the total length of the cord at the bottom of the motion. The final conclusion is that the velocity of the wave can be expressed as √2g(L0+x), taking into account the mass of the block and the cord.
  • #1
norcal36
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Homework Statement
A block with mass M is hanging from a rubber cord, supported such that the cord is unstretched at length L0. The mass of the cord is m which is much smaller than M. The cord has spring constant K. If the block is released: Find the tension, length of cord and speed of wave at the lowest point.
Relevant Equations
V=sq.rt T/K
Sum of the forces
Conservation of energy
Hello again,
I did a sum of the forces at the top and bottom of the motion and got:
Top...mg=ma
Bottom...T-mg=ma
Got T=2mg when I substitute the mg in for ma at bottom.Then I assumed kx=T=2mg so x(length of cord) would equal 2mg/k
Then V of wave would be ...sq.rt2mg/k
Is it that simple? Feel like I'm missing something. Thank you for your help
 
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  • #2
norcal36 said:
Homework Equations:: V=sq.rt T/K
Check your notes. I believe that K here should be replaced by the mass per unit length of the cord.
(You can use the editing toolbar to get symbols such as the square root symbol √ )

I did a sum of the forces at the top and bottom of the motion and got:
Top...mg=ma
Bottom...T-mg=ma
Got T=2mg when I substitute the mg in for ma at bottom.
It appears to me that to get this result you have assumed that the value of ma at the top is the same as the value of ma at the bottom. This might be true. But it needs to be justified.

Another approach for finding the distance x that the mass M falls is to use energy concepts.
 
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  • #3
You are right the equation I used is for another situation. However if I use the conservation of energy I get mgx=1/2kx^2...mg=1/2kx...kx=2mg...kx=T=2mg. Again I am assuming tension will be the equivalent to kx. The x would be the displacement of the "spring" which equals length. However I am still stuck on the velocity of the wave. thanks for your time again.
 
  • #4
norcal36 said:
You are right the equation I used is for another situation. However if I use the conservation of energy I get mgx=1/2kx^2...mg=1/2kx...kx=2mg...kx=T=2mg. Again I am assuming tension will be the equivalent to kx.
OK.
The x would be the displacement of the "spring" which equals length.
x is how much the cord is stretched from its natural length. So, how would you express the total length of the cord when M is at the bottom?

Then, can you find an expression for the mass per unit length of the cord when it is at the bottom?
 
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  • #5
I would express the total length by L0+x...?
Then with that I could use V=√T/m/L equation which then would give me V=√2g(L0+x)...This sound correct?
 
  • #6
norcal36 said:
I would express the total length by L0+x...?
Yes
Then with that I could use V=√T/m/L equation which then would give me V=√2g(L0+x)...This sound correct?
Be sure to distinguish between the mass of the block and the mass of the cord.
 
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FAQ: Is my solution for the Bungee Tension Problem correct?

How do I know if my solution for the Bungee Tension Problem is correct?

The best way to determine if your solution is correct is to compare it with other known solutions and see if they yield the same results. You can also double check your calculations and assumptions to ensure they are accurate.

What factors should I consider when solving the Bungee Tension Problem?

When solving the Bungee Tension Problem, you should consider factors such as the length and elasticity of the bungee cord, the weight of the person jumping, and the height of the jump. These factors will affect the tension on the bungee cord and ultimately determine if the solution is correct.

Are there any common mistakes to avoid when solving the Bungee Tension Problem?

One common mistake is forgetting to account for the weight of the bungee cord itself. This can significantly affect the tension on the cord and lead to an incorrect solution. Other mistakes include miscalculating units or using incorrect formulas.

Can I use a computer program to solve the Bungee Tension Problem?

Yes, computer programs can be used to solve the Bungee Tension Problem. However, it is important to understand the underlying calculations and assumptions being made by the program to ensure the accuracy of the solution.

What other real-life applications can the Bungee Tension Problem be used for?

The Bungee Tension Problem can be applied to various scenarios, such as designing suspension bridges, calculating tension in cables for ski lifts, and determining the force on a bungee jumper during a jump. It can also be used in engineering and physics education as a practical application of mathematical concepts.

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