Question

A 55kg woman bungee jumps from a bridge and reaches a maximum velocity of 24.2m/s. The bridge is 60m above the water. When the woman is at the bottom of her trajectory, she is 9 m above the water.
What is the spring constant of the bungee cord?

Answer 1

Answer on Question #58900-Physics-Quantum Mechanics

A $m = 55kg$ woman bungee jumps from a bridge and reaches a maximum velocity of $V = 24.2m/s$. The bridge is $H = 60m$ above the water. When the woman is at the bottom of her trajectory, she is $h = 9m$ above the water.

What is the spring constant of the bungee cord?

Solution

The maximal kinetic energy is

$$EK_{max} = \frac{mV^2}{2}.$$

The woman reaches a maximum velocity at the equilibrium of two forces:

$$mg = kx \rightarrow x = \frac{mg}{k}$$

At this point

$$E_{total} = \frac{mV^2}{2} + \frac{kx^2}{2} + mg(h_{eq} - h)$$

When the woman is at the bottom of her trajectory

$$E_{total} = \frac{k(x + (h_{eq} - h))^2}{2}$$

When the woman is at the bottom of her trajectory

$$E_{total} = mg(H - h)$$ $$\frac{mV^2}{2} + \frac{kx^2}{2} + mg(h_{eq} - h) = mg(H - h) \rightarrow mg(H - h_{eq}) = \frac{mV^2}{2} + \frac{kx^2}{2} = \frac{mV^2}{2} + \frac{m^2g^2}{2k}$$ $$h_{eq} = H - \frac{V^2}{2g} - \frac{mg}{2k}.$$ $$mg(H - h) = \frac{k(x + (h_{eq} - h))^2}{2}$$ $$mg(H - h) = \frac{k\left(\frac{mg}{k} + \left(H - \frac{V^2}{2g} - \frac{mg}{2k} - h\right)\right)^2}{2}$$ $$2mg(H - h) = k\left(H - h + \frac{mg}{2k} - \frac{V^2}{2g}\right)^2$$ $$\sqrt{2mg(H - h)} = \sqrt{k}\left(H - h - \frac{V^2}{2g}\right) + \frac{1}{\sqrt{k}} \frac{mg}{2}$$ $$k\left(H - h - \frac{V^2}{2g}\right) - \sqrt{k} \sqrt{2mg(H - h)} + \frac{mg}{2} = 0$$ $$k \left(60 - 9 - \frac{24.2^2}{2 \cdot 9.8}\right) - \sqrt{k} \sqrt{2 \cdot 55 \cdot 9.8(60 - 9)} + \frac{55 \cdot 9.8}{2} = 0$$ $$k(21.12) - \sqrt{k}(234.47) + (269.5) = 0$$ $$\sqrt{k} = \frac{234.47 + \sqrt{234.47^2 - 4 \cdot 21.12 \cdot 269.5}}{2 \cdot 21.12} = 9.8$$ $$k = 96 \frac{N}{m}$$

The other root is physically impossible $k = 1.7 \frac{N}{m}$. We can see it when calculate $x$:

$$x = \frac{mg}{k} = \frac{55 \cdot 9.8}{1.7} = 317 \, m.$$

But the height of the bridge is just $60 \, \text{m}$ and the position of equilibrium cannot exceed it.

Answer: $96 \frac{N}{m}$.

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