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, 84 kg. The dean would stand on top of a platform that is 43 m above a 2.0 m deep pool of Jello with a 30 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 30 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 11m, 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? 558 N/m

Explanation / Answer

this is the equilibrium position

ie the net force on him is zero

two forces acting on him

1) his weight --vertically downwards

2) elastic force of bungee card --- vertically upward

both forces arecting on him in air . therefore g = 9.8 m/s2

so that, elastic force of bungee card = his weight = 84*9.8 =823.2 N

                     1/2 kx2 =84*9.8 =823.2 N

                        1/2 k(11)2 =823.2 N

                                  k = 13.6 N/m

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