00:01
So here, for part a, we're going to use equation 811 in order to find the initial elongation, the elongation in the initial situation.
00:10
So we can say that the elongation initial would be equal to the square root of 2 times 1 .44, which would be the elastic potential energy of the spring.
00:22
And then this is going to be divided by 3 ,200 newtons per meter.
00:28
That would essentially be the spring.
00:30
Constant of the spring, and this is equaling 0 .30, whether 0 .030 meters or 3 .0 centimeters.
00:41
We can say that in the next situation, the elongation is only 2 centimeters.
00:48
So now we have less stored energy.
00:51
The change in potential energy would then be equal to one -half times 3 ,200 newtons per meter.
01:03
Multiplied by 0 .020 meters, quantity squared minus the initial potential energy of the spring.
01:13
So minus 1 .44 joules.
01:16
And so the change in potential energy would be equal to negative 0 .80 joules.
01:22
This is your answer for part a.
01:24
For part b, we know that the elastic stored energy for a spring displacement of 0 .022...