Posted by **rei** on Saturday, January 14, 2012 at 9:54pm.

An ore car of mass 41000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 9.6 m lower vertically, is a horizontally situated spring with constant 4.7 × 10^

5 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.

- physics - Energy -
**drwls**, Saturday, January 14, 2012 at 10:06pm
Gravitational potential energy loss = Spring potential energy increase

M g H = (1/2) k X^2

k = 4.7*10^5 N/m

M = 4.1*10^4 kg

H = 9.6 m

You known what g is.

Solve for X

- physics - Energy -
**rei**, Saturday, January 14, 2012 at 10:10pm
Thank you!

- physics - Energy -
**rei**, Saturday, January 14, 2012 at 10:16pm
I got it wrong. I did exactly what you said. Multiplied m times g times h then multiplied by 2. Sq root that number and got 2778.926411 but it was wrong.

- physics - Energy -
**drwls**, Sunday, January 15, 2012 at 12:08am
X = sqrt(2 M g H/k)

= 4.54 m

You forgot to divide by k, the spring constant.

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