Find the launch speed of a ball in a spring mechanism

Homework Statement

I have a question asking me to find the launch speed of a ball (mass 0.39kg) when released by a spring mechanism made of 2 springs each with force constant 25Nm^-2. they are pulled back 12 cm. the ball is initially at rest.

[/B]
v^2=u^2+2as
f=ma
f=kx
E=1/2kx^2
E=1/2mv^2

The Attempt at a Solution

I used f=kx=ma to find the acceleration and then v^2=u^2+2as to find v however I got a different answer (1.92) as the mark scheme which used conservation of energy, elastic potential energy (1/2kx^2)=kinetic energy (1/2mv^2). The answer they got was 1.4Shouldn't I have got the same answer and if not what am I doing wrong and why is it wrong to use SUVAT in this situation?

Related Introductory Physics Homework Help News on Phys.org
BvU
Homework Helper
2019 Award
Hello Izwx,
I used f=kx=ma
What did you take for a ?
Or, equivalently: what for x ?

haruspex
Homework Helper
Gold Member
why is it wrong to use SUVAT in this situation?
SUVAT is only for constant acceleration. As the spring expands the force it applies diminishes, so the acceleration drops.

Hello Izwx,
What did you take for a ?
Or, equivalently: what for x ?
I got 15.38 ms^-2

BvU
Homework Helper
2019 Award
And is that valid for the whole launching interval ?

SUVAT is only for constant acceleration. As the spring expands the force it applies diminishes, so the acceleration drops.
oh I see thank you

And is that valid for the whole launching interval ?
I think thats what someone else was getting at that it wasn't constant acceleration so you can't use Suvat

BvU
Homework Helper
2019 Award
Same way spring energy is 1/2 kx2 and not kx2

(your answer is a factor of $\sqrt 2$ higher than the correct book answer)

Same way spring energy is 1/2 kx2 and not kx2

(your answer is a factor of $\sqrt 2$ higher than the correct book answer)
ah yes that makes a lot of sense thanks a lot