When trying to stop the train, Spiderman applies a 4000 N force over 200 seconds. What is his impulse on the train?

Initial speed of train = 35.76 m/s (80 mph)

               I figured out the speed of the train (which equals 80mph) by viewing the movie. In one of the scenes, it is shown that the train reaches 80mph after Doc Ock desroys the break on the train.

                   1mph = 0.44704mps

                    80(0.44704) = 35.7632

                                      = 35.76m/s

   vf = 0m/s

   vi = 35.76m/s

Mass of train = approximately 137,100 kg (302,254 lbs)  (found in research)

Time taken to stop train = approximately 46 seconds

a = vf - vi

       t

  = 0 - 35.76

         46

  = -.7773913043

  = -.78 m/s^2

(The number is negative since the accelerarion of the train is slowing down)

According to Newton's Second Law, a = f/m. When rearranged, the formula reads f = m(a). As shown in the free body diagram, the weight of the train equals the force due to gravity.

(force = tension)

F =  m(a)

  = 137,100(-.78)

  = -106938 N

106938 N is the calculated tension of Spider-Man's webs

I also calculated the distance it took for Spider-Man to stop the train:

Approximate distance traveled by train:

d = 1/2(a)(t)^2

  = 1/2(-.78)(46)^2

  = 825.24 m

In order for Spider-Man to have been able to completely stop the train before falling off the edge of the tracks, I found that the tension of his webs must have equaled 106938N.

In order to determine the tension of Spider-Man's webs, I needed to know the mass of the train. Since the movie was unclear on what type of trian was used, I researched online to find the mass of the train. Knowing the mass of the train, i was then able to calculate the tension of Spider-Man's webs and the distance the train took to stop.

Explanation:

well clustered together.

points close on a scatter diagram

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