Question
Asked Dec 15, 2019
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A 75 kg student jumps off a bridge with a 12-m-long bungee cord tied to his feet. The massless bungee cord has a spring constant of 430 N/m.

a)    How far below the bridge is the student’s lowest point?

b)   How far below the bridge is the student's resting position after the oscillations have fully stopped?

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Expert Answer

Step 1

(a) The gravitational potential energy is converted into the spring potential energy

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mgh=kar mgh=K(h-1

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Step 2

Substitute the values,

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(75 kg)(9.8 m/s²)(12m)=÷ (430 N/m)(h- (12 m))* (41.02 m²) = (144 m²) +h² - (24 m)h h - (24 m)h + (102.98 m²) = 0 h=18.4 m

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Step 3

(b) When the oscillation is died of, the gravitational force acting on the jumper is equal t...

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F, = F, kAr = mg k(1 – h') = mg mg

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Work,Power and Energy

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