Energy stored in a bow and arrow

[vectordog]

The following was a question from a Work/Energy high school test.
"3. An archer is shooting at a hay bale target. The archer pulls back on his bow string with 500 N of force to draw it back a distance of 80 cm. He releases the string, and the arrow (m= 1 Kg) flies straight to its target (neglect air resistance).
a) How fast is the arrow traveling when it leaves the bow?"
Some students used the work-energy theorem (W=delta KE in this case) and got an answer of 28.3 m/s. However, many students used the info given to determine a spring constant (k) for the bow (k = F/x), and then applied Conservation of Energy (Elastic PE = KE in this case) to the system. This yields an answer of 20 m/s for the arrow.
My question is "Why doesn't the spring constant method work?"

 

[Doc Al]

Some students used the work-energy theorem (W=delta KE in this case) and got an answer of 28.3 m/s.

They messed up.

However, many students used the info given to determine a spring constant (k) for the bow (k = F/x), and then applied Conservation of Energy (Elastic PE = KE in this case) to the system. This yields an answer of 20 m/s for the arrow.

That's fine.

My question is "Why doesn't the spring constant method work?"

Both methods, done correctly, yield the same answer. Note that the spring force varies from 500 N at first to zero at the release. You can't just use W=Fx = 500*0.8.

 

[vectordog]

THANKS!
I feel a little stupid now.