physics????

A archer puts a .3kg arrow into the bowstring. An average force of 201N sis exerted to draw the bowstring 1.3m.

a) if there is no friction, what velocity does the arrow have when it leaves the bow?

When i worked it out using two different ways, the answer is conpletely different. They are supposed to be the same.

here's the first way:

To find the energy store in the bowstring

W=Fx

W= (201N) (1.3m)

W= 261.3J

W=KE

W= 1/2mv^2

261.3J = 1/2 (.3kg)v^2

V= 41.73m/s

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here is another way that i used.

I find the spring constant first

F=kx

201N = k(1.3m)

k= 154.62 N/m

EPE = 1/2kx^2

EPE = 1/2(154.62N/m)(1.3m)^2

EPE = 130.65J

EPE=KE

EPE=1/2mv^2

130.65J = 1/2(.3kg)v^2

v= 29.5m/s

i don't know which way is correct, this really drives me crazy. If the first way is correct, then why not the second way? and if the second way is correct, then why not the first way?

can somebody please explain??? i really need help.

Update:

first of all, thank you for answering.

and second, you didn't really answer my question. i doubt if you read the last paragraph.