Superman must stop a 150 km/h train in 190 m to keep it from hitting a stalled car on the tracks.

a) If the train's mass is 4.0×105 kg, how much force must he exert (find the magnitude)?

b) Compare to the weight of the train (give as %).

c) How much force(in magnitude) does the train exert on Superman?

10pts to the most detailed and CLEAREST answer. If you explain you steps with words as well, it would be a big bonus

## Answers (2)

Okay, we are more than 1/2 way through the semester, I am going to assume you have heard of energy and work by now. (You can do this problem without using energy and work, it just takes a few more steps.)

The energy of the train is all KE, and KE = 0.5 x M x v^2

The force, F, that must do work against the train over the distance, d, will equal the energy of the train.

(The energy of the train is taken from the train by Superman who is doing work, he is applying a force over a distance.)

So, KE = 0.5 x M x v^2 = W = F x d

or 0.5 x (4.0 x 10^5 kg) x (150 km/h x 1000m/km x 1 hr/3600s)^*2 = F x 190 m

3.47 x 10^8 J = F x 190 m

1.83 x 10^6 N = F (well that takes care of part A).

The weight of the train is M x g where g = 9.81 m/s^2 (gravitational acceleration)

So the weight = 4.0 x 10^5 kg x 9.81 m/s^2 = 3.92 x 10^6 N

To calculate a percent (1.83 x 10^6 N / 3.92 x 10^7 N ) x 100 = 46.7% So, superman has

to push with 46.7% of the force it would take to pick up the train!

Finally, according to Newton, if Superman pushes on the train with 1.83 x 10^6 N of force, the train pushes back on the Superman with the same force. So, the train is pushing back with 1.83. x 10^6 N of force.

Now, that problem was not so hard. Once you realize what a physicist would call the work-energy theorem. You can add or subtract energy with work!

I bet you could have answered part b and c after you were helped through part a. Is that not true?

1st step convert DATA to BASIC SI units:

V = 150 km/h = (150)(1000)/3600) m/s = 41.7 m/s

d = 190 m

a) m = 4.0e5 kg

KE of train = 1/2mV² = (0.5)(4.0e5)(41.7)² = 1739(2.0e5) = 3.48e8 J

Work to stop train in distance = d:

Favg x d = 3.48e8

Favg = 3.48e8/1.90e2 = 1.83e6 N ANS a)

b) Wt of train = 4.0e5 x 9.81 = 39.24e5

Favg/Wt =1.83e6/3.92e6 = 0.467 ANS b)

c) force of train on S-man = Favg = 1.83e6 N ANS c)