Work/Energy with spring

Heres the problem

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Heres my work

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First off I want to make sure im understanding the question. Is it asking to find the speed at B after the collar is shot up by releasing the spring that was depressed 1.5 in.?


And if that is correct, im trying to figure how to balance the equations incorporating the depressed spring....

 

I understand that I need to use conservation of energy but the spring is not at rest.....so would this work?

.5kx^2 = .5mv^2 + .5kx^2
(x,1) ( x,2)




and would the velocity be v^2 = a*r

 

ok i calculated the numbers and came up with v = 31.12.....however I noticed when i divide 31.12 by the Weight of 6 oz. .... i come up with 5.18 which is the answer in the back of the book......but that doesnt make sense to me....ounces needs to be converted.....

 

no thats what i meant that it made no sense...im just trying to figure why i came up with 31 and the answer is 5.18 ft/s

 

ok, so than I could come up with another equ.

.5mv^2 = .5mv*2 + mgh

and plug in the V I found on the left side of the equ.

but for finding V on the right side of the equation do I need to take into consideration the radius of curvature, such as tangential or normal components..

 

i realized when I did the first calculation with 1.5 inches for x that I didnt convert to feet.........so i converted that as well as everything else....so i get a V of 2.55 ft/s.....thats the collar leaving the spring


so mass = .012
10 in. = .833 ft
1.5 in, = .125ft

i plugged into the equ.


.5(.012)(2.55)^2 = .5(.012)(V)^2 + .012(32.2)(.833)

.039 = .006v^2 + .321

V is supposed to be 5.18 but its not quite coming out.....