A $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{-}\mathbf{\text{kg}}$subway train is brought to a stop from a speed of$\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{\text{m}}\mathbf{/}\mathbf{\text{s}}$in$\mathbf{0}\mathbf{.}\mathbf{400}\mathbf{\text{m}}$by a large spring bumper at the end of its track. What is the force constant k of the spring?

Force constant: When a spring is stretched by the application of some force, then according to the Hooke’s law the increase in the length of spring from its equilibrium length is directly proportional to the force applied.

Mathematically,

$$\begin{gathered} F\propto x \\ F=kx \end{gathered}$$

Here, k is the proportionality constant known as force constant or spring constant.

When a spring is stretched, the change in potential energy of the spring is,

$\Delta \text{PE}=\frac{1}{2}k{x}^2$

Here, k is the force constant, and x is the displacement$\left(x=0.400\text{m}\right)$.

According to the law of conservation of energy,

$$\begin{gathered} \Delta \text{PE}=\Delta \text{KE} \\ \frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2 \end{gathered}$$

Here, m is the mass of the subway train$\left(m=5.00\times {10}^5\text{kg}\right)$, and v is the initial velocity of the subway train$\left(v=0.500\text{m}/\text{s}\right)$.

The expression for the force constant is,

$k=\frac{m{v}^2}{x^2}$

Putting all known values,

$$\begin{gathered} k=\frac{\left(5.00\times {10}^5\text{kg}\right)\times {\left(0.500\text{m}/\text{s}\right)}^2}{{\left(0.400\text{m}\right)}^2} \\ =781250\text{N}/\text{m} \end{gathered}$$

Therefore, the required force constant of the spring is $781250\text{N}/\text{m}$.