Chapter 4, Problem 5P

Superman must stop a 120-km/h train in 150 m to keep it from hitting a stalled car on the tracks. If the train's mass is 3.6 x 10⁵ kg, how much force must he exert? Compare to the weight of the train (give as %). How much force does the train exert on Superman?

To determine

The force exerted by the Superman and the force that the train exerts on Superman.

Answer to Problem 5P

Solution:

The force exerted by Superman on the train is $1.33\times {10}^5\text{ N}$ and when compared to the weight of the train is $37.8\%$.

The force exerted by the train on the superman is in opposite direction of the force exerts by Superman on the train which is $-1.33\times {10}^5\text{ N}$

Explanation of Solution

The force on an object is calculated by using Newton’s second law of motion as given below:

$F=ma$………… (1)

Here, $m$ is the mass of the object and $a$ is its acceleration

Now, by using Newton’s equation of motion, the expression for the acceleration is given as follows:

$v^2-u^2=2as$………… (2)

Here, $s$ is the distance, $u$ is the final speed of the person and $v$ is its initial speed.

Given:

Mass of the train $m=3.6\times {10}^5\text{ kg}$

Distance $s=150\text{ m}$

The initial speed of the person $u=120\text{ km/h}$

Final speed of the person $v=0\text{ km/h}$

Formula used:

$F=ma$

$v^2-u^2=2as$

Calculation:

Substitute, $m=65\text{ kg}$ and $u=120\text{ km/h}$ in equation (1)so the force exerted by a person is given as below:

  F=ma

    =(65 kg)(-30 g)

    =(65 kg)(-30)(9.8 m/s²)

    =1.9×10⁴N

Now, we have to substitute,, = 33.33 m/s and $v=0\text{ km/h}$ in equation (2)so the distance traveled by a person is given as follows:

                v²-u²=2as

 0-(33.33 m/s²)=2a(150 m)

                       a=3.7 m/s²

Now, substitute $a=3.7{\text{ m/s}}^2$ and $m=3.6\times {10}^5\text{ kg}$ in equation (1)so the force exerted by the superman on the train is given as follows:

 F₁=ma

    =(3.6×10⁵ kg)(3.7 m/s² g)

    =1.33×10⁵N

Now we must compare the force exerts by superman on the train $F_1$ by the weight of the train as given below:

$\left(\frac{ma}{mg}\right)\times 100\%=\frac{a}{g}\times 100\%$

                          $=\frac{3.7}{9.8}\times 100\%$

                          $=37.8\%$

The force exerted by the train on the super is also same as the force $F_1$ i.e. force exerted by Superman on the train but in opposite direction.