When bungee jumping from a high bridge over Victoria Falls, an operator first at

Question

When bungee jumping from a high bridge over Victoria Falls, an operator first attaches an elastic rope to the jumper. The jumper then jumps off the bridge, falling freely until they reach the unstretched length of the rope. Then, the rope begins to stretch and slows the jumper to a stop. The rope pulls the jumper back up, and they oscillate up and down for a while until the operator pulls the jumper back up to the bridge.

The rope is essentially a long spring. Let's label the jumper's mass m, the unstretched length of the rope L0, the height of the bridge above the water H, the elastic (spring) constant of the rope k, and the gravitational field strength g. The jumper's speed at the point where they've fallen the full length of the unstretched rope is a maximum, vmax. For simplicity, we will neglect resistive forces like air drag.

Explanation / Answer

E.

Use law of conservation of energy

Ei=Ef

mgH= ½*mvmax2 + mg(H-L0)

‘m’ cancels from both sides,

gH= ½*vmax2 + mg(H-L0)

Thus vmax is independent of mass ‘m’.

Plugging values,

10*150=1/2* vmax2 +10*(150-11)

Gives,

vmax =14.8 m/s

Equating eqn from B and D we get,

1/2*mvmax2 = 1/2*kd2 –mgd

Plugging values,

½*80*14.8^2=1/2*k*11^2-80*10*11

k=753.42

Thus spring constant is larger in later case.

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