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.(adsbygoogle = window.adsbygoogle || []).push({});

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 im doing the right thing?

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# Homework Help: Mechanical Energy question

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