Posted by **GiGi** on Tuesday, November 11, 2008 at 7:46pm.

A 3.30 kg object is hanging from the end of a vertical spring. The spring constant is 48.0 N/m. The object is pulled 0.200 m downward and released from rest. Complete the table below by calculating the translational kinetic energy, the gravitational potential energy, the elastic potential energy, and the total mechanical energy E for each of the vertical positions indicated. The vertical positions h indicate distances above the point of release, where h = 0.

Find KE, PE-gravity, PE-elastic and E for each h (0, 0.200, 0.400)

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**drwls**, Tuesday, November 11, 2008 at 8:32pm
The spring is already stretched by the weight before is is pulled. That amount of "equilibrium" stretch is

Xe = M g/k = 0.674 m. Then the spring is pulled downward another 0.200 m for a total stretch of X = 0.874 m at the point of release. The kinetic energy is zero at that point, and the spring potential energy is (1/2)kX^2 = 18.33 J . Relative to the the point of release, the position is then h = 0. Call the gravitational potential energy zero at that point also.

The total energy of the system is the spring energy of 18.33 J

At h = 0.2, X = 0.674 m

At h = 0.4, X = 0.474 m

gravitational PE (gPE) = M g h

spring potential energy = (1/2) k X^2

kinetic energy = 18.33 J - KE - gPE

You complete the table

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