Question

A 40,000 kg subway train is brought to a stop from a speed of 0.700 m/s in 0.250 m by a large spring bumper at the end of its track. What is the force constant k of the spring?

Answer 1

Given,

The mass of the train, m=40000 kg

The initial velocity of the train, u=0.700 m/s

The compression in the spring bumper that stopped the train, x=0.250 m

The final velocity of the train, v=0 m/s

From the equation of motion,

`v^2-u^2=2ax`

Where a is the acceleration of the train.

On substituting the known values,

`\begin{gathered} 0-0.700^2=2a*0.250 \\ \Rightarrow a=(-0.700^2)/(2*0.25) \\ =-0.98\text{ m/s}^2 \end{gathered}`

The magnitude of the force applied by the train will be equal to the magnitude of the restoring force of the spring.

Therefore,

`\begin{gathered} m|a|=kx \\ \Rightarrow k=(m|a|)/(x) \end{gathered}`

Where k is the spring constant of the spring.

On substituting the known values,

`\begin{gathered} k=(40000*0.98)/(0.250) \\ =156800\text{ N/m} \end{gathered}`

Therefore the spring constant of the spring is 156800 N/m

Thus the correct answer is option C.