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Find the launch speed of a ball in a spring mechanism

  • #1
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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?
 

Answers and Replies

  • #2
BvU
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Hello Izwx, :welcome:
I used f=kx=ma
What did you take for a ?
Or, equivalently: what for x ?
 
  • #3
haruspex
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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.
 
  • #4
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Hello Izwx, :welcome:
What did you take for a ?
Or, equivalently: what for x ?
I got 15.38 ms^-2
 
  • #5
BvU
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And is that valid for the whole launching interval ?
 
  • #6
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SUVAT is only for constant acceleration. As the spring expands the force it applies diminishes, so the acceleration drops.
oh I see thank you
 
  • #7
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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
 
  • #8
BvU
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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)
 
  • #9
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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
 

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