Maximum acceleration using energy principle

  • Thread starter shehan1
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1. Design a "bungee jump" apparatus for adults. A bungee jumper falls from a high platform with two elastic cords tied to the ankles. The jumper falls freely for a while, with the cords slack. Then the jumper falls an additional distance with the cords increasingly tense. Assume that you have cords that are 14 m long, and that the cords stretch in the jump an additional 22 m for a jumper whose mass is 80 kg, the heaviest adult you will allow to use your bungee jump (heavier customers would hit the ground).

(a) It will help you a great deal in your analysis to make a series of 5 simple diagrams, like a comic strip, showing the platform, the jumper, and the two cords at the following times in the fall and the rebound:
1 while cords are slack (shown here as an example to get you started)
2 when the two cords are just starting to stretch
3 when the two cords are half stretched
4 when the two cords are fully stretched
5 when the two cords are again half stretched, on the way up
On each diagram, draw and label vectors representing the forces acting on the jumper, and the jumper's velocity. Make the relative lengths of the vectors reflect their relative magnitudes.

(b)Focus on this instant of greatest tension and, starting from a fundamental principle, determine the spring stiffness ks for each of the two cords.
ks = 58.31404959 N/m

(c)What is the maximum tension that each one of the two cords must support without breaking? (This tells you what kind of cords you need to buy.)
FT = 1282.909091 N

(d) What is the maximum acceleration |ay| = |dvy/dt| (in "g's") that the jumper experiences? (Note that |dpy/dt| = m|dvy/dt| if v is small compared to c.)


2. Ef = Ei + W
Kf + Uf = Ki + Ui +W
acceleration in m/s^2 divided by 9.8 m/s^2

3. I tried setting 1/2 mv^2 + 1/2 Ksf^2 = mgh If you could help please explain
 

Answers and Replies

  • #2
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Nevermind i figured this one out :p
 

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