Given:
The mass of the subway train is,
[tex]m=5.2\times10^5\text{ kg}[/tex]
The initial speed of the train is,
[tex]v=0.55\text{ m/s}[/tex]
The distance moved by train is,
[tex]x=0.51\text{ m}[/tex]
To find:
the spring constant
Explanation:
The kinetic energy of the train converts into the potential energy of the spring. So we can write,
[tex]\begin{gathered} \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ k=\frac{mv^2}{x^2} \end{gathered}[/tex]
Substituting the values we get,
[tex]\begin{gathered} k=\frac{5.2\times10^5\times0.55^2}{0.51^2} \\ =0.60\times10^6\text{ N/m} \end{gathered}[/tex]
Hence, the spring constant is
[tex]0.60\times10^6\text{ N/m}[/tex]
