Conservation of Energy w/ Tarzan & Jane

In summary, Jane needs to swing across a river filled with crocs in order to rescue Tarzan. However, she must swing into a constant horizontal wind Force (110 N) on a vine (L=40m) that is initially at an angle of (theta=50) with the vertical. The distance of the river she must cross is D=50 m. Jane's minimum speed to just make it to the other side is 6.15 m/s, and once the rescue is complete, Tarzan and Jane must swing back together across the river. Jane's minimum speed to begin their swing is 9.87 m/s.
  • #1
njvc3
4
0
Hello, I was looking for some help. please...

I've been struggling with the following problem:

Jane, whose mass is 50kg, needs to swing across a river filled with crocs in order rescue Tarzan, whose mass is 80kg. However, she must swing into a constant horizontal wind Force (110 N) on a vine (L=40m) that is initially at an angle of (theta=50) with the vertical. and the distance of the river she must cross is D=50 m. (The diagram displays the height Jane is at inititally to be a little higher than the height of the bank Tarzan is at)
a) What is the minimum speed Jane must begin her swing in order to just make it to the other side?
b) Once the rescue is complete, Tarzan and Jane must swing back together across the river. What is the minimum speed they must begin their swing?


I tried to figure out the height by figuring out how much Potential Energy she has.. Change in net Energy is 0; so using the Conservation of Energy equation: delta KE + delta Epot + deltaEwind = 0

figuring the final velocity (that she lands at) is 0 and the change in final height is 0... I'm left with the following equation:
(1/2m*(initial v)^2)+(m*g*delta h initial)-(Force wind*D)

where the only information I don't have is the initial v and delta h initial.


The answers are supposed to be a)6.15m/s & b) 9.87m/s ... but, i can't seem to get it..

Thanks. :smile:
 
Physics news on Phys.org
  • #2
a) Apply conservationa of energy . Use the angles given to calculate the height difference.

BJ
 
  • #3
This is my first time posting on the forums and I'm not a native English speaker, plus I'm unfamiliar with the proper Physics terms, so bear with me.


Jane starts at 50 degrees, and lands at [tex]\alpha = arcsin(D/L - cos40)[/tex] = 28,94406598 degrees (compared to the vertical line), which you can get by applying basic trigonometry. Now you should be able to solve the difference in height.

[tex]W = FL[sin50 + sin \alpha] = 5500 J[/tex] (I went through some integrating but, of course, got the same answer as you'd get with W = FD)
 
Last edited:
  • #4
njvc3 said:
Hello, I was looking for some help. please...

I've been struggling with the following problem:

... (The diagram displays the height Jane is at inititally to be a little higher than the height of the bank Tarzan is at)...

There appears to be some missing information. Is this height difference labeled? Is the vine placed between the river (25 m from each side)? I just love these tarzan and Jane swinging problems. This one appears to be straight forward but the drawing or more clarification would be helpful.
 
  • #5
pete worthington said:
There appears to be some missing information. Is this height difference labeled? Is the vine placed between the river (25 m from each side)? I just love these tarzan and Jane swinging problems. This one appears to be straight forward but the drawing or more clarification would be helpful.

There is enough information given. The position of the vine relative to the river is determined by its length and the initial angle.
 

Attachments

  • TarzanJane.png
    TarzanJane.png
    10.5 KB · Views: 797
  • #6
You shoud use the theorem of kinetic energy:

W=K2-K1=-K1 (K2 is at least 0)
W=W(gr)+W(wind)
=-(U2-U1)-integral(from theta1 to -theta2 which you determine with geometric reasons)[M*dtheta]

U is the potential energy for gravity.
M=F*r*sin(theta+pi/2); integral[M*dtheta]=-F*r*(sin(theta1)+sin(theta2))

They are all simple calculations.
 
  • #7
Thank you all..

Thanks for your insight.. That is exactly what I was not understanding.. I did have to use the angle. Thanks! :biggrin:
 

1. What is the concept of conservation of energy?

The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but can only be transformed from one form to another.

2. How does the story of Tarzan and Jane relate to conservation of energy?

In the story, Tarzan and Jane use their surroundings and their own energy to perform various tasks and survive in the jungle. This showcases the principle of conservation of energy as they are constantly transforming and utilizing energy in different forms.

3. What are some examples of conservation of energy in the story of Tarzan and Jane?

Some examples include Tarzan using his muscles to swing from vines and Jane using the energy from fire to cook food. Both of these actions demonstrate the transformation of energy from one form to another.

4. How does the conservation of energy apply to real-life situations?

The principle of conservation of energy applies to all physical systems and processes in the universe. It can be seen in everyday situations such as a ball rolling down a hill, a car accelerating, or a light bulb producing light.

5. Is it possible to violate the conservation of energy?

No, the conservation of energy is a fundamental law of physics that has been proven through various experiments and observations. It is not possible to violate this principle as it is a fundamental aspect of the universe.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
887
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
11K
  • Introductory Physics Homework Help
Replies
3
Views
5K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
4K
Back
Top