# Elastic Potential Energy Question tough

• cheechnchong
In summary, we are given the mass of a climber, the spring constant of a nylon rope, and the distance he falls before the rope starts to slow him down. Using the energies of kinetic and potential, we can find the distance the rope is stretched when it stops his fall. However, there may be confusion in where to define the potential energy to be zero, and it is important to consider the kinetic and potential energies at three different points - the top of the cliff, where the rope starts to catch him, and where the rope finally stretches to its maximum. By choosing a location to define the potential energy as zero, we can solve for the stretch of the rope using conservation of energy.

#### cheechnchong

Elastic Potential Energy Question! tough

An 86.0 kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of 1.20 x 10^3 N/m. He accidentally slips and falls freely for 0.750 m before the rope runs out of slack. How much is the rope streched when it breaks his fall and momentarily brings him to rest?

This is my approach:

First i found the energies...

KE = 1/2(86)(0) = 0J
PE = (86)(9.8)(0) = 0 J
PE(elastic) = = 1/2ky^2 = 1/2(1.2 x 10^3) (.75)^2 = 338J

THEN, i put The total energy in this equation to solve for elastic distance..

E = KE + PE + PE(elastic)

338J = 1/2(86kg)(0m/s) + (86)(9.8 m/s^2)(.750m) + 1/2(1.2 x 10^3 N/m) (x)

x = 1.62 m (i know I am wrong b/c the book answer is different)

Last edited:
Be careful about where he is when the rope is at equilibrium, and where you defined the potential energy to be zero. I think that's where the issue is...

I'd say really look at the potential and kinetic energy at 3 points -- the top of the cliff, where the rope starts to catch him, and where the rope finally streches to its maximum. The energy of each position should equal energy of each other position.

physics girl phd said:
Be careful about where he is when the rope is at equilibrium, and where you defined the potential energy to be zero. I think that's where the issue is...

I'd say really look at the potential and kinetic energy at 3 points -- the top of the cliff, where the rope starts to catch him, and where the rope finally streches to its maximum. The energy of each position should equal energy of each other position.

hmmmm...so like this?

PE (at the top) = (86)(9.8)(0)
KE = 0
PE (catched) = (86)(9.8)(.750) = 632.1 J
PE (elastic) = 1/2(1.2 x 10^3)(.750)^2 = 338 J<----i think the x-value is

E(total) = 970J

so after that step:

970 = (86)(9.8)(0) + (86)(9.8)(.750) + 1/2(1.2e-3)(x)

^^^i think the confusion part is my prob...so how do i approach that??

Last edited:
You ARE corrent in that you are confused about the x.

I'd choose to define his potential energy to be zero at the place the rope first kicks in (that way you can think of the rope like a normal spring displaced from equilibrium).

Assuming this -- what is his potential energy at the top? What is his kinetic energy at the top?

physics girl phd said:
You ARE corrent in that you are confused about the x.

I'd choose to define his potential energy to be zero at the place the rope first kicks in (that way you can think of the rope like a normal spring displaced from equilibrium).

Assuming this -- what is his potential energy at the top? What is his kinetic energy at the top?

hmmmmm, i know for sure the PE (at the top) is 0 because h = 0. Now for the KE...1/2v(initial)^2 -- that's a toughie! where can i obtain this value with only 2 givens (vfinal = 0 and y=.750m)? is g = 9.8 m/s^2 the third value?

Is he moving at the top?

physics girl phd said:
Is he moving at the top?

no, not at all...does potential energy have the larger value??

so what is his KE at the top?

physics girl phd said:
so what is his KE at the top?

zero lol...??

im really having a hard time figuring the PE elastic though.

Last edited:
I think you need to recognize a few important things.

1) you have THREE points of interest.
a)top of cliff
b) point where spring kicks in
c) point where spring is stretched.

2) You need to find the KE and PE energy at EACH location.
KE (top) +PE (top) = KE (middle) +PE (middle) = KE (bottom) +PE (bottom)

3) YOU get to chose a location where the potential energy is ZERO. You seem to want to do this at the top. I'd personally do it at the middle position to make the rope like a standard "spring" problem -- then finding x is the same as a spring. But ONCE you choose a location, you need to calculate all the potential energies relative to this position, which could even mean that the potential energies are negative (if you chose 0 at the top, and his KE is 0, then you get a positive KE as he falls, balanced out by the negative PE -- conserving energy, make sense?).

.

^I don't even know what the velocity is at the bottom!?? v =0 as well right?

YES!

(the rope stopped him!)

physics girl phd said:
YES!

(the rope stopped him!)

lol ok...this is what i got! but it still doesn't correspond to the book answer

1/2(86)(3.38 m/s) + (86)(9.8)(0) = 1/2(86)(0) + 1/2(1200)x --- i solved from the middle and the bottom

145.3 = 600x

x = .242 m

(sorry if I am not gettin it down...i have a sleeping urge lol)

Last edited:
It might be a square on the velocity if things are just like you typed above(I'm not really checking numbers here, just giving you ideas to think about). If you are getting brain-drain, take a break and look at this thread again tomorrow (It's 2 AM here too!).

physics girl phd said:
It might be a square on the velocity if things are just like you typed above(I'm not really checking numbers here, just giving you ideas to think about). If you are getting brain-drain, take a break and look at this thread again tomorrow (It's 2 AM here too!).

haha ok i guess i will...i'll PM you with anymore trouble i have with this prob! what a beastly prob

cheechnchong said:
An 86.0 kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of 1.20 x 10^3 N/m. He accidentally slips and falls freely for 0.750 m before the rope runs out of slack. How much is the rope streched when it breaks his fall and momentarily brings him to rest?

This is my approach:

First i found the energies...

KE = 1/2(86)(0) = 0J
PE = (86)(9.8)(0) = 0 J
PE(elastic) = = 1/2ky^2 = 1/2(1.2 x 10^3) (.75)^2 = 338J

THEN, i put The total energy in this equation to solve for elastic distance..

E = KE + PE + PE(elastic)

338J = 1/2(86kg)(0m/s) + (86)(9.8 m/s^2)(.750m) + 1/2(1.2 x 10^3 N/m) (x)

x = 1.62 m (i know I am wrong b/c the book answer is different)
The rope is not stretched at all in the beginning. If you choose the initial position to be the zero of potential energy as you have done (which is as good a choice as any), then the total energy is zero. Since you are calling y the amount by which the rope is stretched, then I suggest you call the position of the climber relative to the starting point something else (I will use y'). As the climber falls, y' becomes increasingly negative, reducing the gravitational potential energy (and increasing kinetic energy). When the rope starts to stretch, GPE continues to get lower, but the elastic energy of the rope increases (decreasing kinetic energy).

What is the relationship between y (rope stretch) and y' (position of climber relative to starting point) while the rope is stretching? What is the total energy, and the contributing parts of that total when the climber is again at rest at the bottom of the fall?

## 1. What is elastic potential energy?

Elastic potential energy is the energy stored in an object when it is stretched or compressed. It is a type of potential energy that is stored in the elastic materials, such as rubber bands, springs, and bungee cords.

## 2. How is elastic potential energy calculated?

Elastic potential energy is calculated using the equation E = 1/2kx^2, where E is the elastic potential energy, k is the spring constant, and x is the displacement of the object.

## 3. What are some examples of elastic potential energy?

Some examples of elastic potential energy include a stretched rubber band, a compressed spring, a bent archery bow, and a stretched bungee cord.

## 4. How is elastic potential energy different from other types of potential energy?

Elastic potential energy is different from other types of potential energy because it is specific to elastic materials and is related to their ability to be stretched or compressed. Other types of potential energy, such as gravitational potential energy, are related to an object's position or configuration.

## 5. How is elastic potential energy useful in everyday life?

Elastic potential energy is useful in everyday life in various ways. It is used in sports equipment, such as bows and trampolines, to provide a source of energy for movement. It is also used in everyday objects, such as rubber bands and springs, to store and release energy for various purposes, like holding objects together or providing tension in machines.

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