How Does Mechanical Energy Explain a Bungee Jump?

[tre2k3]

TA Chuck Margraves spends his spare time bungee jumping off the Henley Street bridge. He has a 14 ft long bungee cord (unstretched) that has a spring stiffness of 22 lb/ft. To test his bungee, Chuck ties the bungee cord to the bridge railing, ties the other end to an unnamed EF professor weighing 134 pounds, and then gently nudges the professor off of the bridge.


A. Determine the professor's speed when the bungee cord starts to stretch (ft/sec)
B. Assuming he hasn't hit the water, how much has the bungee chord stretched when professor is at the bottom of the jump? (ft)

The formula I used to get the answers was 1/2mgv^2+mgh+1/2k(x)^2+Win=1/2mv^2+mgh+1/2k(x)^2+Eloss
Since there was no work or Eloss and i set my datum at and inital velocity 0 and my initial x to 14ft so I came up with the final formual: 1/2(22)(14)^2=1/2(134lbs)(v)^2.
Did I use the right approach, and do I need to convert?

For B, I used KE+PE=KE+PE.I set my inital and final velocity 0, my height is the distance fall before the cord stretches + the change in y where the bungee cord go to the lowest point in which i got the formula, 0+mg(14ft+y)=0+1/2k(y)^2

i set it up as a quadratic formula to get the answer, but I was wondering do I need to convert and If I am doing the right thing?

 

[siddharth]

The first part is fairly simple and all we need is energy conservation. That is the potential energy lost = kinetic energy gained

v= \sqrt{2gh}
where h is 14 feet.

For the second part the spring stiffness is mass per unit length? That can't be because then the force would be in a wrong dimension

Assuming the spring constant to be k, then work done by spring in expanding will be 1/2kx^2 (x from equlibirium position, ie 14 feet). This will equal to a further loss in P.E as the velocity of the professor at the bottom is 0. Just check on the units of the spring constant though.

 

[tre2k3]

im not getting the correct answer for B. I did it by just the numbers they gave me and tried to do it by converting to regular units such as N/M or m/s^2

 

[tre2k3]

yeah, lb/ft had me going cause i needed to cancel out some other units so I just did a conversion and I got the right answer