Finding the Right Spring Force: A Tale of Error and Redemption

[pandatime]

Homework Statement

Let's say we have a bungee cord where from 0<x<4.88 the spring constant (k) is 204 N/m, and when x>4.88 the spring constant is 111 N/m. What is the tension in the cord when the stretch is 16.7m?

Relevant Equations

$F_s = -kx$

What I've done so far is find the spring force through
$F_s = -kx$
$F_s = -111*16.7$
$= -1853.7N$

My conclusion was that since this is the spring force, the tension force must be just the negative of that so $1853.7N$ because the net force has to balance out, but I am horribly horribly wrong.

the answer should be $2.32kN$, which is a few hundred off from my answer. I'm thinking that there is maybe a force I'm missing in my calculations? Or does this problem have to do with potential energy of any sorts?

 

[Lnewqban]

Why did you use the 111 value?

 

[haruspex]

Homework Statement:: Let's say we have a bungee cord where from 0<x<4.88 the spring constant (k) is 204 N/m, and when x>4.88 the spring constant is 111 N/m. What is the tension in the cord when the stretch is 16.7m?
Relevant Equations:: $F_s = -kx$

What I've done so far is find the spring force through
$F_s = -kx$
$F_s = -111*16.7$
$= -1853.7N$

My conclusion was that since this is the spring force, the tension force must be just the negative of that so $1853.7N$ because the net force has to balance out, but I am horribly horribly wrong.

the answer should be $2.32kN$, which is a few hundred off from my answer. I'm thinking that there is maybe a force I'm missing in my calculations? Or does this problem have to do with potential energy of any sorts?

The usual definition of a spring constant, applied tension/total extension, assumes that it is actually constant. How are we to interpret it if it is extension-dependent?
If we continue to use applied tension/total extension something weird must happen at the boundary. As we take it beyond the critical length, the tension suddenly halves.
More likely, they intend the definition $\frac{\Delta T}{\Delta x}$. See if that gets the official answer.

 

[pandatime]

The usual definition of a spring constant, applied tension/total extension, assumes that it is actually constant. How are we to interpret it if it is extension-dependent?
If we continue to use applied tension/total extension something weird must happen at the boundary. As we take it beyond the critical length, the tension suddenly halves.
More likely, they intend the definition $\frac{\Delta T}{\Delta x}$. See if that gets the official answer.

the definition of $\frac{\Delta T}{\Delta x}$?

I'm not sure I follow, wouldn't that just lead to the same formula of

$k = \frac{\Delta T}{\Delta x}$
$111 N/m = \frac{\Delta T}{16.7m}$ which was what I did in my attempted answer above which was unfortunately not the answer

Correct me if I'm wrong that that's what you meant

 

[pandatime]

Why did you use the 111 value?

I used it because the problem says that once $x > 4.88$ then the spring constant $k$ is $111 N/m$

 

[Lnewqban]

I used it because the problem says that once $x > 4.88$ then the spring constant $k$ is $111 N/m$

Could it be possible that there is a different force while the bungee stretches from zero to 4.88 meters?

 

[haruspex]

the definition of $\frac{\Delta T}{\Delta x}$?

I'm not sure I follow, wouldn't that just lead to the same formula of

$k = \frac{\Delta T}{\Delta x}$
$111 N/m = \frac{\Delta T}{16.7m}$ which was what I did in my attempted answer above

Correct me if I'm wrong that that's what you meant

The $k = \frac{\Delta T}{\Delta x}$ form means that if at some extension x the tension is T(x) and the 'constant' is k(x) then at a small further extension $x+\Delta x$ the tension is $T(x)+k(x)\Delta x$.
It might have been clearer had I written $k = \frac{d T}{d x}$.

 

[pandatime]

Could it be possible that there is a different force while the bungee stretches from zero to 4.88 meters?

I'm not quite sure. Would that force matter though since the problem asks for the tension at 16.7 metres?

The $k = \frac{\Delta T}{\Delta x}$ form means that if at some extension x the tension is T(x) and the 'constant' is k(x) then at a small further extension $x+\Delta x$ the tension is $T(x)+k(x)\Delta x$.
It might have been clearer had I written $k = \frac{d T}{d x}$.

i'm not sure if 'constant' implies that k is changing throughout the problem even past 4.88 metres?

$k = \frac{d T}{d x}$ how do we find $dT$ if we don't know any value of the tension anywhere along the bungee cord?

 

[Lnewqban]

I'm not quite sure. Would that force matter though since the problem asks for the tension at 16.7 metres?

Yes, it does.
The force of the second stage (16.7-4.88 meters) should be added to the force of the first stage (4.88-0 meters).

 

[haruspex]

$k = \frac{d T}{d x}$ how do we find $dT$ if we don't know any value of the tension anywhere along the bungee cord?

You might be confused about the meaning of x there. I am using it to refer to the amount by which the cord has been extended, not to a position along the cord. While 0<x<4.88 the spring constant (k=dT/dx) is 204 N/m, and when x>4.88 the spring constant is 111 N/m.
Sketch that as a curve of T against x.

 

[Delta2]

@haruspex wants to say that the curve T(x) must be continuous everywhere, especially at the point x=4.88. If you haven't been introduced to the concept of continuity of a curve/function then I don't know how we supposed to explain this to you.

 

[pandatime]

Yes, it does.
The force of the second stage (16.7-4.88 meters) should be added to the force of the first stage (4.88-0 meters).

ok so i did $\int_0^{4.88} 204 dx + \int_{4.88}^{16.7} 111 dx$ and I got 2307N? it's still like 200N off the answer but am I on the right track?

 

[Delta2]

That's what I get too, 2307N.

 

[pandatime]

That's what I get too, 2307N.

unfortunately the answer is 2.32kN so there must be something missing... it's quite close tho...

I am glad we are on the same page though i felt quite lost

 

[Delta2]

Its just 13N difference though not 200N...

 

[pandatime]

Its just 13N difference though not 200N...

omg sorry it is like 5am here right now I'm tripping HAHA thank you, so I guess this is the answer even if it is slightly off? I just worry because this is just practice for me to understand concepts, I don't really care about the answer because it's not for marks but I just don't want it to be the wrong answer which would make it practice for nothing

thank you all so much for helping

 

[Delta2]

Well if the official answer was 2.31kN (rounded up version of 2307) then I would be 100% sure that we ve found the correct solution, however the official answer is 2.32kN which leaves some small doubt.