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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.

Answer :

Answer

given,

mass of jumper = 70 Kg

height when he is at rest = 28 m

time = 42 s

Time period, T = [tex]\dfrac{42}{8}[/tex]

                      T = 5.25 s

now,

[tex]T = 2\pi\sqrt{\dfrac{m}{k}}[/tex]

[tex]5.25= 2\pi\sqrt{\dfrac{70}{k}}[/tex]

[tex]\dfrac{70}{k}=0.698[/tex]

k = 100.26 N/m

extension of the chord

 [tex]x = \dfrac{mg}{k}[/tex]

 [tex]x = \dfrac{70\times 9.8}{100.26}[/tex]

        x = 6.84 m

un stretched length of the bungee cord

L = 28 - 6.84

L = 21.16 m

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