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**Finding the spring constant of a spring...**

## Homework Statement

A mass is hung from a vertical spring. The mass is .200kg, and the spring stretches .086m. Find the spring constant.

## Homework Equations

ΔE=ΔKE+ΔUg+ΔUs

## The Attempt at a Solution

I first broke up the equation, to see which values would go to zero. I know ΔKE drops out, because the initial and final velocities would be zero. Eventually, I was left with this...

mgΔy=(1/2)K(x^2)

Then, I solved for K and got...

K=(2mgΔy)/(x^2)

Putting the numbers in, I get...

K=(2*.200kg*9.80m/s^2*.086m)/(.086m^2)

The final answer I got was approximately 45.6 N/M. I just want to make sure this is correct, along with one more quick thing. The next part of the problem asks, "If the mass is now lifted .086m back to the spring's unstretched position and released from rest, how far will it drop before recoiling back?"

I'm assuming that if this is a closed system, wouldn't the mass just drop down the same distance over and over again? (oscillate the same?) Or do I need to calculate the new distance?

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