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.
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.
Yes, that answer of 4.9 x 10^-2 J is correct for the kinetic energy of the pellet as it leaves the spring.
To calculate the potential energy stored in an ideal spring, you can use the formula PE = 1/2 kx^2, where PE is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
In the first scenario, you have a force of 10N holding the spring in compression, and the spring constant is 20 N/m. Using the formula PE = 1/2 kx^2, you can substitute the values: PE = 1/2 (20 N/m) (x)^2.
Since the force applied to the spring is in the opposite direction of the displacement, the displacement will be negative. Therefore, you should get PE = 1/2 (20 N/m) (-10N/20 N/m)^2 = 1/2 (20 N/m)(0.5)^2 = 1/2 (20 N/m) (0.25) = 2.5J. So the correct answer is 2.5J, not 10J.
In the second scenario, you have a spring with a constant of 20N/m and it is initially compressed by 7.0cm. You want to find the kinetic energy of the pellet as it leaves the spring.
To find the potential energy stored in the spring, you can again use the formula PE = 1/2 kx^2. Substituting the values, you get PE = 1/2 (20 N/m) (0.07m)^2 = 1/2 (20 N/m) (0.0049m^2) = 0.049J or 4.9 x 10^-2 J.
However, the potential energy only tells us the amount of stored energy in the spring. To find the kinetic energy of the pellet as it leaves the spring, you need to consider the conservation of energy. At the point of release, all of the potential energy is converted into kinetic energy.
So the kinetic energy of the pellet as it leaves the spring is equal to the potential energy stored in the spring, which is 0.049J or 4.9 x 10^-2 J. Hence, your revised answer of 4.9 x 10^-2 J is correct.