How far back up will the guy recoil?

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Homework Help Overview

The problem involves a bungee jumper who descends from a bridge attached to a bungee cord. The jumper's mass, the length of the bungee cord, and the spring constant are provided. The discussion focuses on determining how far the jumper descends and how far he will recoil back up after the jump, considering the effects of air friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of energy principles to analyze the motion of the bungee jumper. There are attempts to set up equations relating to gravitational potential energy, elastic potential energy, and kinetic energy. Questions arise regarding the interpretation of variables and the setup of the equations.

Discussion Status

Some participants have provided equations to describe the energy transformations during the jump and recoil. Clarifications are being sought on the meanings of specific terms and variables used in the equations. The discussion is ongoing, with participants exploring different aspects of the problem without reaching a consensus.

Contextual Notes

Participants are working under the assumption that air friction only affects the jumper on the way up, and there is a specific mention of the average work done by friction during the return trip.

ussjt
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Could someone help me with thsi problem...

A 65-kg bungee jumper jumps straight down from the tiop of a bridge. He is tied to a 15-meter-long bungee cord that has an effective spring constant of 400 N/m. (a) How far from the top of the bridge will the bungee-jumper descend(Ignore air friction). (b) Assume that air friction only acts on the guy on the way up. If the average work due to friction for the return trip is 175 joules/meter, how far back up will the guy recoil?

I would just like to know the set up.
 
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Use energies.

[tex]W_{R - mg - F_{ele}} = \Delta E_m = \Delta E_p + \Delta E_{ele} + \Delta E_k[/tex]

On his way down the equation would look like this:

[tex]0 = mg(h_2 - h_1) + \frac{1}{2}k(x_2^2 - x_1^2) + \frac{1}{2}m(v_2^2 - v_1^2)[/tex]
Where:
[tex]h_2[/tex], [tex]h_1[/tex] - The final and initial heights above your potential reference point
[tex]x_2[/tex], [tex]x_1[/tex] - The final and initial elongation of the spring
[tex]v_2[/tex], [tex]v_1[/tex] - The final and initial speeds of the diver (both zero)
What you need to do is solve this equation for ([tex]h_2 - h_1[/tex]) which is distance of the jumper from the top of the bridge.

On the way it would look like this:

[tex]W_{friction} = -f(h_3 - h_2) = mg(h_3 - h_2) + \frac{1}{2}k(x_3^2 - x_2^2) + \frac{1}{2}m(v_3^2 - v_2^2)[/tex]
Again, solve for ([tex]h_3 - h_2[/tex]).

Hope this helps.
 
Could you explain what everything stands for in the first equation you gave me...I seem to use different abbriviations.

Other than that, thanks a lot.
 
[tex]W_{R - mg - F_{ele}} = \Delta E_m = \Delta E_p + \Delta E_{ele} + \Delta E_k[/tex]

Means:

The work done by non-preserving forces in our system equals to the change in mechanical energy of the system, which is the sum of the changes in gravitational potential energy, elastic potential energy and kinetic energy of the system.

(I'm sorry, I should have used "els" instead of "ele" since the latter is usually used when dealing with electricity.)
 
thank you a lot.
 

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