# How do I find the maximum extension?

• Strange07
In summary: The equation for the maximum extension is: E = mgh - TKEwhereE = the energy (in Newtons)m = the mass (in kilograms)g = the gravitational force (in Newtons)h = the height (in meters)So, the equation tells us that the energy of the system is equal to the mass plus the gravitational force, minus the potential energy of the system. In this case, the potential energy is the tension in the rope.In summary, the man jumps and reaches a maximum extension, which is equal to the tension in the rope. The energy of the system is equal to the mass plus the gravitational force
Strange07
Homework Statement
If a static load of mass 20kg extends an elastic rope by 0.8m, what is the maximum extension that a falling man of mass 70kg who jumps off a high cliff tied to this elastic rope would produce, given that the rope is 10m long?
Relevant Equations
F=kx, hence x should be F/k. Since k is constant in both cases, F(man)/F(load)=extension(man)/extension(load)
I found the ratio between the weight exerted by the 70kg man to the weight exerted by the 20kg load, and equated it to the ratio of maximum extension by the 70kg man to the extension by the load
Is this a right step to use, because the book didn’t give any answer nor solution which can help me know whether I’m doing the right thing, and if not, what’s the correct solution please

When the man finally comes to rest (after bouncing up and down a few times) the tension in the rope will equal his weight. Your method gives the extension for this equilibrium position.

But the maximum extension correponds to the lowest point reached; this will be below the equilibrium position. So a different method is needed.

Hint. What are the energy changes from the initial position to the lowest position?

EDIT - accidentally pasted item removed.

Last edited:
Lnewqban
https://stokes.byu.edu/teaching_resources/bungee.html

Steve4Physics
Lnewqban said:
That's an interesting graph - there is a point where the three types of energy are equal. Not sure if that's generally true or just an accident due to the way the drawing has been done.

Lnewqban
Steve4Physics said:
That's an interesting graph - there is a point where the three types of energy are equal. Not sure if that's generally true or just an accident due to the way the drawing has been done.
It seems to depend on the slack length divided by the maximum drop being ##\frac{3-\sqrt 3}6##.

Steve4Physics
Hi @Strange07. Are you still there? Do you still need help?

Steve4Physics said:
Hi @Strange07. Are you still there? Do you still need help?
Yeah
I’m quite confused. So if the method I used is for the equilibrium position, then how do you determine the lowest point from the question given please?

Strange07 said:
Yeah
I’m quite confused. So if the method I used is for the equilibrium position, then how do you determine the lowest point from the question given please?
We're not going to tell you! (That's how it works here.) But we might be able to guide you.

Can you give the formulae for:
gravitational potential energy
elastic potential energy
(and while you're at it) kinetic energy
?

Steve4Physics said:
We're not going to tell you! (That's how it works here.) But we might be able to guide you.

Can you give the formulae for:
gravitational potential energy
elastic potential energy
(and while you're at it) kinetic energy
?
Yhyg
I know the GPE is mgh, TKE is 0.5mv^2 and EPE is 0.5kx^2, where x is the extension

Strange07 said:
Yhyg
I know the GPE is mgh, TKE is 0.5mv^2 and EPE is 0.5kx^2, where x is the extension
I'll assume 'Yhyg' is not excessively offensive.
Kinetic energy is usually abbreviated to just 'KE' not 'TKE'.

Here are some steps for you to work through - do as many as you can and post your answers

Let's call the maximum rope extension 'x'.

Q1. What is the height-change when the mass reaches the lowest postion (an expression which includes x)?
Q2. How much potential energy has the mass lost from the start point to the lowest position (an expression which includes x)?
Q3. What is the weight of the mass (in Newtons)?
Q4. What is the elastic rope's spring constant (k)?
Q5. What is the elastic potential energy stored in the rope when the mass is at the lowest point (an expression which includes x)?
Q6. Can you use the law of conservation of energy and the answers to Q2 and Q5 to write an equation?
Q7. Can you now solve the equation? (I think it's a quadratic equation, so you need to know how to sove quadratic equations).

Edit: typo's corrected.

In post #10, Q3 should be
Q3. What is the weight of the 20kg mass (in Newtons)?

Steve4Physics said:
I'll assume 'Yhyg' is not excessively offensive.
"Yes, here you go "?

haruspex said:
"Yes, here you go "?
Svp
(Sounds very plausible.)

The man jumps. He is initially in free fall until he falls the length of the rope. At this point, there is an additional upward force which eventually brings him to a stop. Note, at this point (the length of the rope) he is still accelerating downward. As a side note, if he was shot downword out of a cannon he would have had an initial velocity, and it would take additional extension of the rope to stop him. Maximum extension occurs when he stops going downward.

One could calculate with forces, but that gets messy. A better way is to do it with conservation of energy. There are three energies in this problem: gravitational, elastic and kinetic. Conservation of energy means that their sum is a constant, C.

What are the energies when the man jumps? This allows you to calculate the constant C.

What are the energies at the point of maximum extension? This will allow you to write an equation for the maximum extension.

Steve4Physics said:
I'll assume 'Yhyg' is not excessively offensive.
Kinetic energy is usually abbreviated to just 'KE' not 'TKE'.

Here are some steps for you to work through - do as many as you can and post your answers

Let's call the maximum rope extension 'x'.

Q1. What is the height-change when the mass reaches the lowest postion (an expression which includes x)?
Q2. How much potential energy has the mass lost from the start point to the lowest position (an expression which includes x)?
Q3. What is the weight of the mass (in Newtons)?
Q4. What is the elastic rope's spring constant (k)?
Q5. What is the elastic potential energy stored in the rope when the mass is at the lowest point (an expression which includes x)?
Q6. Can you use the law of conservation of energy and the answers to Q2 and Q5 to write an equation?
Q7. Can you now solve the equation? (I think it's a quadratic equation, so you need to know how to sove quadratic equations).

Edit: typo's corrected.
caz said:
The man jumps. He is initially in free fall until he falls the length of the rope. At this point, there is an additional upward force which eventually brings him to a stop. Note, at this point (the length of the rope) he is still accelerating downward. As a side note, if he was shot downword out of a cannon he would have had an initial velocity, and it would take additional extension of the rope to stop him. Maximum extension occurs when he stops going downward.

One could calculate with forces, but that gets messy. A better way is to do it with conservation of energy. There are three energies in this problem: gravitational, elastic and kinetic. Conservation of energy means that their sum is a constant, C.

What are the energies when the man jumps? This allows you to calculate the constant C.

What are the energies at the point of maximum extension? This will allow you to write an equation for the maximum extension.
Steve4Physics said:
I'll assume 'Yhyg' is not excessively offensive.
Kinetic energy is usually abbreviated to just 'KE' not 'TKE'.

Here are some steps for you to work through - do as many as you can and post your answers

Let's call the maximum rope extension 'x'.

Q1. What is the height-change when the mass reaches the lowest postion (an expression which includes x)?
Q2. How much potential energy has the mass lost from the start point to the lowest position (an expression which includes x)?
Q3. What is the weight of the mass (in Newtons)?
Q4. What is the elastic rope's spring constant (k)?
Q5. What is the elastic potential energy stored in the rope when the mass is at the lowest point (an expression which includes x)?
Q6. Can you use the law of conservation of energy and the answers to Q2 and Q5 to write an equation?
Q7. Can you now solve the equation? (I think it's a quadratic equation, so you need to know how to sove quadratic equations).

Edit: typo's corrected.
I get the concept now
So if I equate the initial GPE to the final EPE, my equation would be mg(10+x)=1/2(k)(x)^2??

Steve4Physics
Strange07 said:
I get the concept now
So if I equate the initial GPE to the final EPE, my equation would be mg(10+x)=1/2(k)(x)^2??
Yes.

## 1. How do I calculate the maximum extension of a spring?

The maximum extension of a spring can be calculated using the formula x = F/k, where x is the maximum extension, F is the applied force, and k is the spring constant.

## 2. What is the significance of finding the maximum extension of a spring?

Finding the maximum extension of a spring is important in understanding its behavior and determining its potential uses. It can also help in predicting the spring's response to different forces and designing efficient spring systems.

## 3. How do I find the spring constant to calculate the maximum extension?

The spring constant can be found by dividing the applied force by the corresponding extension of the spring. This can be done by conducting experiments and plotting a graph of force versus extension, where the slope of the line represents the spring constant.

## 4. Can the maximum extension of a spring be exceeded?

Yes, the maximum extension of a spring can be exceeded if a force greater than its elastic limit is applied. This can permanently deform the spring and affect its future behavior.

## 5. How does the material of a spring affect its maximum extension?

The material of a spring can affect its maximum extension by influencing its stiffness and strength. Different materials have different spring constants, and some materials may have a higher elastic limit, allowing for a greater maximum extension before permanent deformation occurs.

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