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e-lub [12.9K]
4 years ago
9

A person of mass m will bungee-jump from a bridge over a river where the height from the bridge to the river is h. The bungee co

rd has an un-stretched length of 23h and when it stretches beyond its equilibrium length it behaves as a spring with spring constant k. Find an expression for the minimum spring constant k that will stop the person just before hitting the river (ignore the height of the person), in terms of m, g, and h.
Physics
1 answer:
Likurg_2 [28]4 years ago
5 0

Answer:k_{min}=\frac{18mg}{h}

Explanation:

Given

mass of person is m

Distance between bridge and river is h

chord has an un-stretched length of \frac{2h}{3}

Let spring constant be k

Person will just stop before hitting the river

Conserve energy i.e. Potential Energy of Person is converted in to elastic energy of chord

mgh=\frac{kx^2}{2}

x=h-\frac{2h}{3}=\frac{h}{3}

mgh=\frac{kh^2}{18}

k=\frac{18mg}{h}

Thus k_{min}=\frac{18mg}{h}

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