Bungee Jumping Energy (with drag)

[akgtdoskce]

Homework Statement

A 102kg man bungee jumps with a 20ft (~6m) cord (k=167 N). Air resistance is 12N.
1. What should be the minimum height of the platform the man jumps off?
2. What is the maximum velocity he will reach?
3. How high will he bounce up the first time?

Homework Equations

Uelastic_initial+ KEinitial + Ugravity = Uelastic_final + KEfinal + Ugravity
Elastic Potential Energy= 1/2kx^2
Kinetic Energy= 1/2mv^2
Gravitational Potential Energy= mgh
mgh = (1/2)k(h-L)^2, where L is unstretched distance

The Attempt at a Solution

I've seen a lot of threads on similar bungee jumping problems, but I'm still not clear how the 12N air resistance would factor in. I know W=Fd so the air resistance would be multiplied by displacement, but beyond that I'm not sure where to add/subtract when applying conservation of energy.
Am I right in thinking that the equation mgh = (1/2)k(h-6)^2 could be used if air resistance is negligible? If so, how could I adjust for the 12N air resistance?
(Also, any additional help on the three questions would be much appreciated... I feel like the correct thinking to get the answers is way beyond me)

 

[akgtdoskce]

Update: Okay, so here's what I attempted to do. Could anyone let me know if this thinking/my answers make sense?to find height: mgh = (1/2)k(h-6.1)^2+W_drag
h= 22.37m.

to find height bounced up to:
mgh=.5k(22.37-6.1)^2-12h
h=21.388 m

to find velocity:
1/2mv^2+W_drag=mgh
.5*102*v^2+12*6.1=9.8*6.1*102... since height here would just be unstretched length of cord, right?
v=10.8685 m/s

Any feedback would be much appreciated!

 

[haruspex]

mgh=.5k(22.37-6.1)^2-12h
h=21.388 m

I get a slightly larger number (less than 22).

height here would just be unstretched length of cord,

Will the jumper stop getting faster as soon as the cord is taut?

v=10.8685 m/s

Looks like you made a decimal place error before square rooting.

 

[akgtdoskce]

I get a slightly larger number (less than 22).

Oops, I actually have 21.838 written down. Was it just a typo or should I completely redo the calculation?

Will the jumper stop getting faster as soon as the cord is taut?

I thought maximum velocity would be at this point because kinetic energy starts being transferred to elastic energy as the cord then stretches.

Looks like you made a decimal place error before square rooting.

I'm doing sqrt (118.1247059) to get the 10.8685 answer...can't quite locate my mistake.

 

[haruspex]

Oops, I actually have 21.838 written down.

That's what I get.

I thought maximum velocity would be at this point because kinetic energy starts being transferred to elastic energy as the cord then stretches.

Yes, but gravity is still doing work. How do you normally find the maximum of a continuous variable?

I'm doing sqrt (118.1247059) to get the 10.8685 answer...can't quite locate my mistake.

No, my mistake - sorry. But the answer is wrong as noted just above.

 

[akgtdoskce]

Yes, but gravity is still doing work. How do you normally find the maximum of a continuous variable?

I have no idea... I've never heard of a continuous variable before and a quick search doesn't really return how you would find the maximum.
Oh wait, is this integrals? If so, it's way beyond the math my class level should capable of.
Would it make sense to set KE+W_drag+GPE equal to the initial GPE?
51v²+73.2+(102*9.8*(22.37-6.1))=22361.052 ? Edit: never mind, this just gives the same answer.

 

[haruspex]

I have no idea... I've never heard of a continuous variable before and a quick search doesn't really return how you would find the maximum.
Oh wait, is this integrals?

I was thinking in terms of calculus, yes, but differentiation. Anyway, we don't really need calculus here.
If an object accelerates up to some speed, then stops getting faster, what can you say about the acceleration now?

 

[akgtdoskce]

If an object accelerates up to some speed, then stops getting faster, what can you say about the acceleration now?

Well if its velocity is now constant, acceleration would be 0. But if the object is slowing down then acceleration is negative.

 

[haruspex]

Well if its velocity is now constant, acceleration would be 0. But if the object is slowing down then acceleration is negative.

Right, so what is the acceleration at the moment of maximum speed?

 

[akgtdoskce]

Right, so what is the acceleration at the moment of maximum speed?

Acceleration would be 0. I'm not exactly sure where you're going with this though.

 

[haruspex]

Acceleration would be 0. I'm not exactly sure where you're going with this though.

So think about forces when the acceleration is zero.

 

[akgtdoskce]

So think about forces when the acceleration is zero.

Well when acceleration is zero there's no net force. However there's still gravity acting on it along with elastic potential energy...? So these would all be equal?

 

[haruspex]

Well when acceleration is zero there's no net force. However there's still gravity acting on it along with elastic potential energy...? So these would all be equal?

"elastic potential energy" is not a force. Other than that, you're on the right lines.

 

[akgtdoskce]

"elastic potential energy" is not a force. Other than that, you're on the right lines.

So force of the "spring" instead? F=-kx 167*x =mg
x=999.6/167= 5.99? I"m confused as to what I'm solving for at this point? Is it displacement/height?

 

[haruspex]

So force of the "spring" instead? F=-kx 167*x =mg

You left out drag.

x=999.6/167= 5.99? I"m confused as to what I'm solving for at this point? Is it displacement/height?

Well, what does x represent in F=-kx?

 

[akgtdoskce]

You left out drag.

F_spring-W_drag=mg
I'm always confused as to whether I should add or subtract drag.
167x-12x=mg
x=6.45 m

Well, what does x represent in F=-kx?

Well x represents displacement from equilibrium, so it's the extra distance that the cord has stretched beyond the 6.1m?

Using this I get velocity to be about 15.59 m/s, unless I"m still missing something.
Edit: also getting 11.17 if I use 6.45m as height.

 

[haruspex]

F_spring-W_drag=mg
I'm always confused as to whether I should add or subtract drag.

At max velocity, which way is the jumper moving, so which way is drag? Is that the same way as mg or opposite?

167x-12x=mg

That's not right. What would the dimension of 12x be?

Well x represents displacement from equilibrium, so it's the extra distance that the cord has stretched beyond the 6.1m?

Yes.

Using this I get velocity to be about 15.59 m/s, unless I"m still missing something.

Much closer.

 

[akgtdoskce]

At max velocity, which way is the jumper moving, so which way is drag? Is that the same way as mg or opposite?

Drag would be in opposite direction of mg, so it would be addition then? I'm not very good at envisioning these things.

That's not right. What would the dimension of 12x be?

Should it just be N then, and not N m?

Is 15.25 m/s any closer? 167x+12=mg, x=5.9

 

[haruspex]

Is 15.25 m/s any closer? 167x+12=mg, x=5.9

Right equation and number for x, but not what I get for v. How did you calculate v? Are you forgetting drag again?

 

[akgtdoskce]

Right equation and number for x, but not what I get for v. How did you calculate v? Are you forgetting drag again?

KE+Wdrag=GPE
.5mv^2+12(12.01)=102*9.8*12.01
v^2=232.5701176
am I using the wrong value of h or rounding prematurely?

 

[haruspex]

KE+Wdrag=GPE
.5mv^2+12(12.01)=102*9.8*12.01
v^2=232.5701176
am I using the wrong value of h or rounding prematurely?

No, it's the potential energy in the cord you've forgotten.

 

[akgtdoskce]

No, it's the potential energy in the cord you've forgotten.

KE+Wdrag+EPE+GPE=GPEinitial
.5mv^2+144.12+(.5*167*144.2401)+(102*9.8*12.01)=102*9.8*22.37
I'm getting a square root of a negative again.. what am I doing wrong?

Edit: for the elastic potential energy, if I use 5.91 instead of 12.01, I get 11.96 m/s.

 

[haruspex]

+(.5*167*144.2401)

Explain how you get the numbers in there.

 

[akgtdoskce]

Explain how you get the numbers in there.

Just realized, should the 144.2401 (12.01^2) be just 5.91^2 instead?
In that case I get 11.96 m/s.

 

[haruspex]

Just realized, should the 144.2401 (12.01^2) be just 5.91^2 instead?
In that case I get 11.96 m/s.

Looks like you have all the equations right now, but I get 13.24 m/s.

 

[akgtdoskce]

Looks like you have all the equations right now, but I get 13.24 m/s.

51v²+144.12+(.5*167*34.9281)+(102*9.8*12.01)=102*9.8*22.37
51v²+144.12+(2916.49635)+(12005.196)=22361.052
sqrt (143.0439147)= 11.96
I guess I still have some numbers wrong in there then?

 

[haruspex]

Looks like you have all the equations right now, but I get 13.24 m/s.

On second thoughts, this is wrong:

...+(102*9.8*12.01)=102*9.8*22.37

That would mean the lost gravitational PE is 102*9.8*22.37 - 102*9.8*12.01, but it isn't. It's 102*9.8*12.01.

 

[akgtdoskce]

That would mean the lost gravitational PE is 102*9.8*22.37 - 102*9.8*12.01, but it isn't. It's 102*9.8*12.01.

So in essence it's just
KE+Wdrag+EPE=GPE
Okay, cool. I get the 13.24 m/s as well.
Thank you for all your help! Not only in answering this question (finally) but also with those concepts that my physics teacher failed to properly teach...

 

[haruspex]

So in essence it's just
KE+Wdrag+EPE=GPE
Okay, cool. I get the 13.24 m/s as well.
Thank you for all your help! Not only in answering this question (finally) but also with those concepts that my physics teacher failed to properly teach...

Phew! You're welcome.