Question

and rolls downhill on tracks from a mine. At the end of the tracks, 21 m lower vertically, is a horizontally situated spring with constant 3.3 × 105 N/m. The acceleration of gravity is 9.8 m/s 2 . Ignore friction. How much is the spring compressed in stopping the ore car?

Answer 1

The spring is compressed 7.15 meters.

Step-by-step explanation:

Let us assume that the mass of the car, m = 41000 kg

Position of the tracks, h = 21 m

Spring constant of the spring,
`k=3.3* 10^5\ N/m`acceleration due to gravity,
`g=9.8\ m/s^2`

Let the spring is compressed by a distance of x in stopping the ore car. In this case, the potential energy is balanced by the increase in spring potential energy such that :

`mgh=(1)/(2)kx^2\\\\x=\sqrt{(2mgh)/(k)} \\\\x=\sqrt{(2* 41000* 9.8* 21)/(3.3* 10^5)} \\\\x=7.15\ m`

So, the spring is compressed 7.15 meters.