Question

A 5.2 × 105 kg subway train is brought to a stop from a speed of 0.55 m/s in 0.51 m by a large spring bumper at the end of its track. What is the force constant k of the spring in N/m?

Answer 1

Given:

The mass of the subway train is,

$m=5.2\times10^5\text{ kg}$

The initial speed of the train is,

$v=0.55\text{ m/s}$

The distance moved by train is,

$x=0.51\text{ m}$

To find:

the spring constant

Explanation:

The kinetic energy of the train converts into the potential energy of the spring. So we can write,

$\begin{gathered} \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ k=\frac{mv^2}{x^2} \end{gathered}$

Substituting the values we get,

$\begin{gathered} k=\frac{5.2\times10^5\times0.55^2}{0.51^2} \\ =0.60\times10^6\text{ N/m} \end{gathered}$

Hence, the spring constant is

$0.60\times10^6\text{ N/m}$