Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: How far back up will the guy recoil?

  1. Mar 31, 2004 #1
    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.
    Last edited: Mar 31, 2004
  2. jcsd
  3. Mar 31, 2004 #2
    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]
    [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.
  4. Mar 31, 2004 #3
    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.
  5. Mar 31, 2004 #4
    [tex]W_{R - mg - F_{ele}} = \Delta E_m = \Delta E_p + \Delta E_{ele} + \Delta E_k[/tex]


    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.)
  6. Mar 31, 2004 #5
    thank you a lot.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook