Arrow is short vetically upwards (Force/Springs)

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Homework Statement



A 110 g arrow is shot vertically from a bow whose effective spring constant is 500 N/m

If the bow is drawn 55 cm before shooting the arrow, to what height does the arrow rise?

Homework Equations



F=ma
U=kx^2 - for Spring (bow)
F=-kx

KE=1/2 mv^2
W=fs

The Attempt at a Solution



I found by using:
F=500 * .55
= 275N

The arrow moves upwards at 275N

It therefore accelerates at:
F=ma
a= F/m
= 275/.110

Gravity is assumed to be -9.8.

Thanks for any help
 
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KICKERMAN360 said:

Homework Statement



A 110 g arrow is shot vertically from a bow whose effective spring constant is 500 N/m

If the bow is drawn 55 cm before shooting the arrow, to what height does the arrow rise?

Homework Equations



F=ma
U=kx^2 - for Spring (bow)
F=-kx

KE=1/2 mv^2
W=fs

The Attempt at a Solution



I found by using:
F=500 * .55
= 275N

The arrow moves upwards at 275N

It therefore accelerates at:
F=ma
a= F/m
= 275/.110

Gravity is assumed to be -9.8.

Thanks for any help
That arrow does not move upwards with 275 Newtons. The bow puts 275 Newtons of upward force on it...

So what is the total acceleration on the arrow after you factor in gravity and compute the acceleration from the spring (275/.110)?
 
I figured that much out but could you explain further. Is it simply like vectors? Where I just minus 9.8 from 2500 and then use 2491.2 as the overall acceleration upwards?