Posted by **Diane** on Sunday, November 5, 2006 at 12:38pm.

A force of 10N holds an ideals spring with 20-N/m spring constant in compression. The potential energy stored in the spring is:

I obtained the answer of 10J. I got thatk= -F/x --> the potential energy of the spring = 1/2kx^2. 10J= 1/2(20N/m)(X)^2. Is that right, i'm not sure.

An ideal spring is used to fire a 1.5g pellet horizontally. The spring has a constant of 20N/m and is intially compressed by 7.0cm. The kinetic energy of the pellet as it leaves the spring:

mass= .015kg

x= .07m

k=20N./m

This one I am unsure about the steps. I have the answer as 4.9 X 10^2 J.

A force of 10n

x=F/k=1/2 m

PE= 1/2 k x^2=1/2 *20*1/4

This is not 10 joules.

Energy in gun spring= 1/2 k x^2

= 1/2 * 20 * (.07)^2

Your answer is off considerably, I get much less than one joule, not 490J.

I'm sorry I had gotten an answer of 4.9 X10^-2. Does that sound better?

sounds exact.

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