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:
This one I am unsure about the steps. I have the answer as 4.9 X 10^2 J.
A force of 10n
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?
physics (please check work) - Henry, Friday, December 25, 2015 at 10:11pm
F = k*d = 10 N.
20*d = 10.
d = 0.5 m. = Distance compressed.
PE = 0.5F * d = 0.5*10 * 0.5 = 2.5 J.