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faust18 [17]
2 years ago
12

A bungee jumper with mass 70.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting

a low point eight more times in 42.0 s. He finally comes to rest 28.0 m below the level of the bridge. Calculate the spring stiffness constant and the unstretched length of the bungee cord.
Physics
1 answer:
Alina [70]2 years ago
6 0

Answer

given,

mass of jumper = 70 Kg

height when he is at rest = 28 m

time = 42 s

Time period, T = \dfrac{42}{8}

                      T = 5.25 s

now,

T = 2\pi\sqrt{\dfrac{m}{k}}

5.25= 2\pi\sqrt{\dfrac{70}{k}}

\dfrac{70}{k}=0.698

k = 100.26 N/m

extension of the chord

 x = \dfrac{mg}{k}

 x = \dfrac{70\times 9.8}{100.26}

        x = 6.84 m

un stretched length of the bungee cord

L = 28 - 6.84

L = 21.16 m

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