# Energy Conservation

In the opening scene of the James Bond film Golden Eye (1995), Bond jumps off a dam with a bungee cord that allows him to slow to a standstill and then enter the bad guys layer. The dam is supposed to be in Russia but is actually the Verzasca Dam in Switzerland which is 220m high.

What is the spring constant of the bungee cord?

Solution: This stunt was done for real and the dam used is in Switzerland. You can go today and make the jump yourself. The dam has a height of 220m. At the bottom before Bond shoots his grapple his velocity has slowed to zero. That means his original potential energy has now been entirely converted to the potential energy of the bungee cord which acts just like a spring. $mgh = \frac{1}{2}kx^2,$

where x is the distance the bungee cord has stretched. If we assume the bungee cord is 80m long then $x = 220m – 80m = 140m$ where we have neglected the fact that Bond is a couple of meters from the bottom when he shoots his grapple. Note that the bungee cord is stretched to 2.75 times its original length, which is fairly typical for bungee cords. If we take Bonds mass to be 80kg, we then have $k = {2mgh \over{x^2}} = {2(80kg)(9.8m/s^2)(220m)\over{(140m)^2}} = 17.6 N/m.$

The upward force on Bond when he reaches the bottom is $F= kx = (17.6N/m)(140m) = 2,464$ N.

This force would give an acceleration of $a = F/m = 2464N/80kg = 30.8 m/s^2$

or 3.14g which is also typical in bungee jumping.

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