Question

A 3.12×10^​5 ​​kg subway train is brought to a stop from a speed of 2.65 miles per hour in 0.453 m by a large spring bumper at the end of its track. What is the force constant k of the spring?

Answer 1

Answer:

Force constant, $k=2.13\times 10^6\ N/m$

Explanation:

It is given that,

Mass of the subway train, $m=3.12\times 10^5\ kg$

Speed of the train, v = 2.65 mph = 1.184 m/s

Distance, x = 0.453 m

The energy stored in the spring is balanced by the kinetic energy of the train such that,

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

k is the spring constant of the spring

$mv^2=kx^2$

$k=\dfrac{mv^2}{x^2}$

$k=\dfrac{3.12\times 10^5\ kg\times (1.184\ m/s)^2}{(0.453\ m)^2}$

k = 2131383.47 N/m

$k=2.13\times 10^6\ N/m$

So, the force constant of the spring is $2.13\times 10^6\ N/m$. Hence, this is the required solution.