Question

A 4.00 ✕ 105 kg subway train is brought to a stop from a speed of 0.500 m/s in 0.600 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

291.67 `N/m`

Explanation:

Here in this case,the kinetic energy of the bus got stored as elastic potential energy of the spring as a result of stopping it.
So,if the spring gets compressed by a distance of `x` and it's force constant is `k`,
its elastic potential energy stored will be `(1/2)k(x^2)`
And if the bus of mass `m` was moving with a speed of `v` ,then its kinetic energy was `(1/2)m(v^2)`
So,equating both, we get, `k`=291.67 `N/m`

Answer 2

The spring constant is `=2.78*10^6Nm^-1`

Explanation:

The mass of the train is `m=4*10^5kg`

The speed of the train is `v=0.5ms^-1`

The kinetic energy of the train is

`KE=1/2mv^2`

`=1/2*4*10^5*(0.5)^2J`

`=0.5*10^5J`

The energy stored in the spring is

`E=1/2kx^2`

The spring constant is `=k`

The compression of the spring is

`x=0.6m`

As,

`"Energy stored in the spring"="Kinetic energy"`

`1/2kx^2=0.5*10^5`

`k=10^5/(x^2)=10^5/(0.6)^2=2.78*10^6Nm^-1`

Answer 3

`sf(k=1.39xx10^(5)color(white)(x)"N/m")`

Explanation:

Firstly we can calculate the deceleration of the train:

`sf(v^2=u^2+2as)`

`:.` `sf(0=0.5^2+(2xxaxx0.6))`

`sf(1.2a=-0.25)`

`sf(a=-0.25/1.2=-0.208color(white)(x)"m/s"^2)`

We can use Newton's Law to find the force needed to stop the train:

`sf(F=ma)`

`sf(F=4.00xx10^(5)xx(-0.208)=-8.333xx10^(4)color(white)(x)N)`

This force will be provided by the buffer for which we use Hooke's Law:

`sf(F=-kx)`

`sf(k=-F/x)`

`sf(k=(8.333xx10^(4))/0.6color(white)(x)"N/m")`

`sf(k=1.39xx10^(5)color(white)(x)"N/m")`