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

Force constant,
`k=2.13* 10^6\ N/m`

Step-by-step explanation:

It is given that,

Mass of the subway train,
`m=3.12* 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,

`(1)/(2)mv^2=(1)/(2)kx^2`

k is the spring constant of the spring

`mv^2=kx^2`

`k=(mv^2)/(x^2)`

`k=(3.12* 10^5\ kg* (1.184\ m/s)^2)/((0.453\ m)^2)`

k = 2131383.47 N/m

`k=2.13* 10^6\ N/m`

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