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An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vertically, is a horizontally situated spring with constant 4.2 × 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 in units of m.

Respuesta :

Answer:

The compression in the spring is 5.88 meters.                

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, [tex]k=4.2\times 10^5\ N/m[/tex]

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

[tex]mgh=\dfrac{1}{2}kx^2[/tex]

x is the compression in spring

[tex]x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m[/tex]

So, the compression in the spring is 5.88 meters.                                                                                                                  

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