Find the launch speed of a ball in a spring mechanism

[Izwx2000]

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.

Homework Equations

[/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?

 

[BvU]

Hello Izwx,

I used f=kx=ma

What did you take for a ?
Or, equivalently: what for x ?

 

[haruspex]

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.

 

[Izwx2000]

Hello Izwx,
What did you take for a ?
Or, equivalently: what for x ?

I got 15.38 ms^-2

 

[BvU]

And is that valid for the whole launching interval ?

 

[Izwx2000]

SUVAT is only for constant acceleration. As the spring expands the force it applies diminishes, so the acceleration drops.

oh I see thank you

 

[Izwx2000]

And is that valid for the whole launching interval ?

I think that's what someone else was getting at that it wasn't constant acceleration so you can't use Suvat

 

[BvU]

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)

 

[Izwx2000]

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