Posted by **Elizabeth** on Saturday, December 1, 2012 at 7:20pm.

A 0.63-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.13 to 0.22 m (relative to its unstrained length), the speed of the sphere decreases from 5.95 to 4.75 m/s. What is the spring constant of the spring? (The sphere hangs from the bottom of the spring.)

Answer in N/m

- PLEASE HELP!! PHYSICS -
**drwls**, Saturday, December 1, 2012 at 7:46pm
The increase in potential energy in stretching from 0.13 to 0.22 m equals the decrease in kinetic energy of the metal sphere. That will let you solve for the spring constant k.

k [(.22)^2 -(.13)^2] = (m/2)[(5.95)^2 - (4.75)^2]

You know the mass m, so solve for k.,

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