# Work

In the following scene from the film Spider-Man 2 (2004) the villain, Dr. Octopus, breaks the controls on an elevated train with many passengers on board. Spider-Man attempts to stop the train before it reaches the end of the tracks.

How much work does Spider-Man do in stopping the train?

Solution: The force Spider-Man exerts is in the opposite direction of the trains motion so the work done will be W=Fd with no extra term for the dot product. From the film it looks once Spider-Man really starts slowing the train down with the webbing he has about 5 blocks to stop the train. A block is about 1/4 of a mile so the total distance d=(5)(0.25mi)(1000m/0.62mi)=400m. We don’t know anything about how Spider-Man’s webbing works so to calculate the force we will need to determine the deceleration of the train and then use Newton’s second law. We can see in the film that the speedometer on the train is at 80mph or v=(80mph)(1000m/0.62mi)(1hr/3600s)=35.8m/s. We can look up the mass of a typical subway car and find it is 38,000kg. It looks like the train has 8 cars so the mass would be M=8(38,000kg) = 304,000kg. If we assume the deceleration is constant we can use our equations for constant acceleration. $v^2 = v^2_0 + 2ad$ $a = -{v_0^2\over{2d}} = -{(35.8 m/s)^2\over{2(400m)}}= -1.6 m/s^2$

so from Newton’s second law the net force is $F= ma = (304,000kg)(1.6m/s^2) = -486,400N$

so now we can calculate the total work done as $W = Fd = (-486,400N)(400m) = -194,560,000J$

This is negative because the work done causes the train to lose energy. To get a sense of how much energy this is a McDonalds Big Mac is 2,300kJ. So to replenish the energy Spider-Man used stopping the train would take eating (194,560,000/2,300,000) = 84.6 Big Macs.

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