As part of a fundraiser, you want the new dean to bungee jump from a crane, and

Question

As part of a fundraiser, you want the new dean to bungee jump from a crane, and must convince him that it is safe for a person of his mass, 69 kg. The dean would stand on top of a platform that is 42 m above a 2.5 m deep pool of Jello with a 26 m long bungee cord attached to his ankles. He would then fall off and straight down, ending up hanging upside-down by his ankles. This first 26 m of the fall before the cord begins to stretch would be free fall. As the bungee cord stretches, it exerts a force with the same properties as the force exerted by a spring. You must determine the elastic constant of the cord so that it stretches only 14 m, which will keep the 2 m tall dean's head just out of the Jello. To simplify your calculation, you decide to assume that the dean's center of mass is at the middle of his body. What must be the spring constant of the bungee cord?

Explanation / Answer

From the question.

Mass of dean (M)= 69kg

as it is free fall, from conservation of energy,

velocity attained before cord streches,

(Including deans centre of mass 1m, the body travels by 26 + 1= 27m before streching)

mgh= 1/2 mv2

69 * 10* 27= 1/2 *69* v2

v=23.2 m/s

now the cord starts streching

Elastic energy of cord is 1/2 k x2

according to conservation of energy during streching

mgh + 1/2 mv2 = 1/2 k x2

69 * 10 * 15 + 1/2 69*23.22 = 1/2 k 142

9660 + 18569.3 = 98k

k=288 N/m

so, the spring constant of the bungee cord is 288 N/m.

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