contestada

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?

Respuesta :

Answer:

The spring is compressed 7.15 meters.                      

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, [tex]k=3.3\times 10^5\ N/m[/tex]acceleration due to gravity, [tex]g=9.8\ m/s^2[/tex]

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 :

[tex]mgh=\dfrac{1}{2}kx^2\\\\x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 41000\times 9.8\times 21}{3.3\times 10^5}} \\\\x=7.15\ m[/tex]

So, the spring is compressed 7.15 meters.