# Chapter 4 - Dynamics: Newton's Laws of Motion - Problems - Page 101: 5

Superman must exert a force of $-1.3\times 10^6 ~N$ on the train. This force is 37% of the weight of the train. The train exerts a force of $1.3\times 10^6 ~N$ on Superman.

#### Work Step by Step

$v_0 = (120 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 33 ~m/s$ We can use this initial velocity to find the acceleration. $a = \frac{v^2-v_0^2}{2x} = \frac{0 - (33 ~m/s)^2}{(2)(150 ~m)} = -3.6 ~m/s^2$ Also, we can use the acceleration of $-3.6 ~m/s^2$ to find the force that Superman must exert on the train: $F = ma = (3.6 \times 10^5 ~kg)(-3.6 ~m/s^2) = -1.3\times 10^6 ~N$ Superman must exert a force of $-1.3\times 10^6 ~N$ on the train. By Newton's third law, the train exerts an equal and opposite force of $1.3\times 10^6 ~N$ on Superman. We can compare this force to the weight of the train. $\frac{1.3\times 10^6 ~N}{(3.6 \times 10^5 ~kg)(9.80 ~m/s^2)}\times 100\% \approx 37\%$

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