# Conservation of Mechanical Energy question

• mighty2000
In summary, the student must stop by the time the bungee cord has stretched 7 meters. The bungee cord has a gravitational potential energy of 71.38 kg and a elastic potential energy of 49 kg. The student's kinetic energy is 7700 J and his speed is 14.683 m/s when he falls from the balloon.
mighty2000
Hey yall, I have a question that is totally stumping me, and I have no idea where to start. I would appreciate it if someone could help me on it. Thanks in advance.

M2k

Here is the question:

In the dangerous sport of bungee-jumping, a daring student jumps from a balloon with a specially designed elastic cord attached to his ankles. The unstretched length of the cord is 25.0 m, the student weighs 700 N, and the balloon is 36.0 m above the river below. Assuming that Hooke's law describes the cord, calculate the required force constant if the student is to stop safely 4.0 m above the river.

What I have is below.

Thanks

Last edited:
Welcome to the forums,

I can't give away the answer... forum rules, but I can give you a few hints.

What is the force provided by the bungee cord?

What types of energy are you converting from and to?

Check in your book how springs work in conservation problems, give the problem a shot, and if you're still stuck, show us what you've got and we'll help you out.

solution so far

Sorry about that, I do have somewhat of a start. I know that hooke's law states that |Fs|= -kx and k is the force constant and x is the distance from x=0

therefore

|Fs|=kd=mg or k=mg/d

700*9.8/36m-4m

k=214.375 N/m

am I even close on this one?

Thanks bro

M2k

maybe

Ok, so I convert Newtons from 700 to lbs by multiplying by 0.224809 and then I get from lbs to kilos I multiply lbs by 0.453592 and I end up with 71.38 Kg.

so

if my initial formula was correct then:

|Fs|=kd=mg or k=mg/d

71.38*9.8/36m-4m = 21.86 N/m constant force?

Does this sound right?

Originally posted by mighty2000
Ok, so I convert Newtons from 700 to lbs by multiplying by 0.224809 and then I get from lbs to kilos I multiply lbs by 0.453592 and I end up with 71.38 Kg.

|Fs|=kd=mg or k=mg/d

71.38*9.8/36m-4m = 21.86 N/m constant force?

Does this sound right?

Solving this problem with energy:

What you want to do is convert the one form of energy to the other.

You have a gravitational potential energy: m*g*h

and you will have a elastic potential energy: [inte] Fspring ds

The gravitational energy is no problem to find. You have the starting height, you have the ending height, you have the weight.

The spring is a bit trickier. First off, as you noted, the force of the spring is k*d, where d is the stretched distance. That means that the force gets bigger as the spring stretches. You'll need to integrate WRT distance.

Also note, that the spring does not start stretching at the beginning of the jump; it starts 25m down.

Yep, energy is the way to go (as you presumably knew since you labeled this "conservation of mechanical energy").

The student free-falls for 25 meters at a constant 9.8 m/sec^2 so you can find his speed at the end of that. His weight is 700 N so you can find his mass and calculate the kinetic energy. In order to stop 4 meters above the river, he must stop by the time the cord has stretched 7 meters. Force is "kx" so the work done by the cord is
integral (x=0 to 7) kx dx= (49/2)k. The work done to stop him must be equal to his kinetic energy (conservation of mechanical energy!).

question

But it asks for the required force constant, do I find that unsing the work-energy theorem? I just solve for k?

Thanks for the input gents.

hopefully I will have it solved today.

I will post my solution here.

Last edited:
Ok I think I got this one going now. Please tell me if I am getting this.

I know that KE=Wcord (conservation of mechanical energy)
thus
E=K+U
Ki+Ui=Kf+Uf

= 0+mgh=1/2mv^2+mgy
**I may be wrong here**
** I need to find the velocity after 25 m**
= vf^2=2g(h-y)
= Vf=sqrt(2*9.8(36-25)) = 14.683 m/s

KE=1/2mv^2
=.5*71.428*14.683^2 = 7700 J
F=kx
Wcord is integral 0 to 7 of kx dx
=1/2kx^2]0 to 7 = k(24.5)

from my first statement:
KE=Wcord
7700=k(24.5)
k=314.286 N/m

?
correct??
?

Thanks all

Originally posted by mighty2000
= 0+mgh=1/2mv^2+mgy
**I may be wrong here**
** I need to find the velocity after 25 m**
= vf^2=2g(h-y)
= Vf=sqrt(2*9.8(36-25)) = 14.683 m/s

First off, that two doesn't go there. gh-gy = g(h-y)

EDIT: Duh. Nevermind. I got it.

Second, why are you solving for velocity in the first place? Energy is conserved, remember. The gravitational potential energy lost is exactly equal to the kinetic energy gained. You can take the number from m*g*[del]h or w*[del]h and set that equal to the spring energy.

Also, the guy is in free fall the entire way down. It looks like you solved for the kinetic energy gained just for the last 9 meters. Remember, when the spring is expanding, the guy is still losing gravitational potential energy, and it's still equal to m*g*[del]h

F=kx
Wcord is integral 0 to 7 of kx dx
=1/2kx^2]0 to 7 = k(24.5)

This looks good

from my first statement:
KE=Wcord
7700=k(24.5)
k=314.286 N/m

It will be correct once you get the right value for the dissipated energy in there.

In this problem, solving for kinetic energy is completely unneccessary. GPE lost = KE gained, so since GPE is easier to calculate from the given information, don't use KE at all!

He falls a total of 32 meters. Whatever else is happening to him in the last 7 meters doesn't matter. He still fell 32 meters. That is the amount of energy that his bungee needs to catch.

Last edited:
He falls a total of 32 meters. Whatever else is happening to him in the last 7 meters doesn't matter. He still fell 32 meters. That is the amount of energy that his bungee needs to catch.

So in the end I will have

(24.5)k=mgh is that what you are saying?

k=700*32/24.5

This gives me a force constant value of 914.216 N/m

warmer?

btw, thanks for your help on this one enigma.

Looks good to me.

Do you understand why?

Last edited:
I do now, I have read and re-read that section of my textbook and everything that you said helped me too. This is what I got from our discussions:

Initial Kinetic energy is 0 in a free fall and Potential energy is equal to the earth(gravity)-mass of object system. This is found from the products of the mass of the object, force of acceleration of gravity and the change in free fall height. I know that the energy is conserved in the cord therefore the work done by the cord is the same as the potential energy lost.

The cord had to do work on the person 25 meters into the freefall and that is when the energy was transferred from the earth-object system into the work of the cord.

Anyway thanks for the enormous effort you put into this forum bro, I appreciate it.

peace out

M2k

Guys... How about solving this using only mechanics?

After falling for 25 meters (the length of the spring unstretched), the guy will have a speed of -sqrt(500) m/s = -22.36 m/s. If we now try to make the distance function where d(0)=0, it will be:

d = v[initial]t + 0.5at^2
= -22.36t + 0.5at^2

If -kd is the force excerted by the spring, then...

a = g - kd/m
= -10 - kd/70

d = -22.36t + (-10 - kd/70)0.5t^2
= -22.36t - 5t^2 - kdt^2/140

d + kdt^2/140 = -22.36t - 5t^2
(1 + kt^2/140)d = -22.36t - 5t^2

d = (-22.36t - 5t^2)/(1 + kt^2/140)

This is just a multivariable function with k and t as the "free" variables. The speed function is (by differentiating in respect to t):

v = [(-22.36 - 10t)(1 + kt^2/140)-(kt/70)(-22.36t - 5t^2)]/(1 + kt^2/140)^2

We now need to find a value of k so that there exist a time T where the height will be -7 (meaning 4 meters above water) and the velocity 0. I'm not really in multivariable calc yet so can someone help me in finishing it, giving hints, or (if it's very hard) checking it with monsters like maple?

Thanks a lot!

Well, actually the rest doesn't really require calculus. It's just really complex to solve (maple or mathematica, anyone?)...

Last edited by a moderator:

## What is the law of conservation of mechanical energy?

The law of conservation of mechanical energy states that the total amount of mechanical energy in a closed system remains constant over time, as long as there are no external forces acting on the system. This means that the sum of kinetic energy and potential energy must remain constant.

## What is an example of the conservation of mechanical energy?

An example of the conservation of mechanical energy is a pendulum swinging back and forth. At the highest point of the swing, the pendulum has maximum potential energy. As it falls, this potential energy is converted into kinetic energy. At the lowest point of the swing, the pendulum has maximum kinetic energy. As it swings back up, this kinetic energy is converted back into potential energy. The total amount of mechanical energy (kinetic + potential) remains constant throughout the motion.

## How does friction affect the conservation of mechanical energy?

Friction is a force that opposes motion and converts some of the mechanical energy into heat energy. This means that in a system with friction, the total amount of mechanical energy will decrease over time. However, the law of conservation of energy still holds true, as the lost mechanical energy is converted into another form of energy (heat).

## Why is the conservation of mechanical energy important?

The conservation of mechanical energy is important because it is a fundamental law of physics that helps us understand and predict the behavior of physical systems. It also allows us to analyze and design machines and devices based on the principle of energy conservation.

## What happens when there are external forces acting on a system in terms of the conservation of mechanical energy?

When external forces are acting on a system, the law of conservation of mechanical energy no longer applies. This is because the external forces are doing work on the system, changing the amount of mechanical energy. However, the law of conservation of energy still holds true, as the total amount of energy in the system (including non-mechanical forms) remains constant.

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