Potential energy stored in spring = (1/2)k d^2= 44.5 d^2
That will be the kinetic energy the block has departing the spring
KE =(1/2)mv^2 = 44.5 d^2
I assume that A to B is this next stretch between the spring and the bottom of the ferris wheelie.
normal force on track = m g
friction force = .34 m g
work done by friction = .34 m g (2.5)
= .85 m g
so kinetic energy at bottom of loop = 44.5 d^2 - .85 m g
Loss of energy going up loop = m g h = 3 m g
Now at the top of the loop for zero force on track:
m v^2/r = m g
(1/2) m v^2 = m g r = (1/2)1.5 m g
Ke at top is therefore .75 m g
So the total kinetic energy at the bottom of the track must be 3.75 m g
3.75 m g = 44.5 d^2 - .85 m g
44.5 d^2 = 2.90 m g
d^2 = (2.90/44.5)(.5)(9.81)
d = .565
again, I appreciate your work but site doesn't accept this answer. I don't know why this is occurring for all the problems that I'm working on.
Beats me, but check my arithmetic carefully.
i tried to insert .6447721757 and the site says round-off error.
Damon, I think you made an arithmetic error here: 3.75 m g = 44.5 d^2 - .85 m g
44.5 d^2 = 2.90 m g
It should be 44.5d^2 = 4.6 m g
Then solve from there, I made that change and got the right answer.
Is this right? I have only one chance to insert the answer so i want to make sure.
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