This is why Spidey stretches in the morning

Going superHERO!

WEll as it was a constantly accelerated system I used $v=u+at$ equation .. where $v=0$ , $u=27 m/s$ , $t=60 seconds$

So a comes out to be $9/20 m/s^2$ .

As the train is of 5 cars and each car of mass $40000$ kg , so the total mass is $\Rightarrow$
$200000$ kg .

Now .... $Force = M \cdot a$

$\Rightarrow$ $F = 200000kg \cdot 9/20 m/s^2$

$\Rightarrow$ $F =\boxed{90000 N}$

So our SuperHERO has to have a force of $\boxed{90000 N}$ to stop the train. Watch out the Vedio .

Assuming $\displaystyle a$ is constant (it is given),
$\text{ Force} = M \frac{\Delta V}{\Delta t}$

$F = 5\times 40000 \text{kg}\times \frac{27\frac{\text{m}}{\text{sec}}}{60 \text{sec}}$

$F = \boxed{90000N}$

After seeing the video : I think thats what Tobey's expressions were when he came to know that Andrew Garfield replaced him (just kidding)

First notice that a standard New York City subway car full of terrified passengers has a mass of roughly $40000$ kg. And Spiderman stops five cars. So the total mass $m$ Spiderman has to stop is equal to $40000\times 5$ kg or $200000$ kg.

Spiderman stops the car in exactly $1$ minute or $60$ seconds And he makes the car go from $27$ m/s to $0$ m/s in that time.

So, the magnitude of acceleration $a$ caused on the car is equal to $\dfrac{27}{60}$ $\text{m/s}^2$ .

Now use $F=ma$ .

Plug everything in you're going to get $F=90000$ newtons.

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I feel like I have solved this problem before.

To the staff, are you working on the Brilliant archive?

Since the train is a five car, the total mass of the train would be 40000 x 5 = 200000 kg. By the formula for momentum, p = mv, the momentum of the train is then 200000 x 27 = 5400000 Ns. Then, using the formula for impulse, Impulse = Fs, the force applied by Spiderman is 5400000/60 = 90000 N.

I just went with impulse here:

$F\cdot t = \Delta p$

$F=m\cdot a$ , and the mass is given as $4 \times 10^{4}kg$ . $\Delta p = p_{f} - p_{0}$ and since we know spidey stops the train, the final momentum is simply 0. We can easy calculate the initial momentum: $p_{o} = mv_{o} = 4 \times 10^{4}kg \cdot 27m/s = 5.4 \times 10^{6}kgm/s$ , so the change in momentum is simply $-5.4 \times 10^{6}kgm/s$ .

Now we can calculate the acceleration using $a = \frac{\Delta p}{m\cdot t}$ . Plugging in numbers we arrive at: a = $\frac{-5.4 \times 10^{6}kgm/s}{2.0 \times 10^{5}kg\cdot 60s} = -0.45m/s^{2}$

Now we can solve for the magnitude of the force required: $\left \| F \right \|=ma = 2.0 \times 10^{5}kg \cdot 0.45m/s^{2} = 9 \times 10^{4}N$

Definitely simpler ways to get at the answer, but it seemed like a fun impulse problem. Hope that was useful. Cheers!