Find the launch speed of a ball in a spring mechanism

In summary, the conversation discusses the calculation of the launch speed of a ball released by a spring mechanism. Different methods are used, including using F=ma and conservation of energy, with the latter providing the correct answer. The incorrect use of SUVAT is also mentioned and the difference between spring energy and force is clarified.
  • #1
Izwx2000
5
0

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?
 
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  • #2
Hello Izwx, :welcome:
Izwx2000 said:
I used f=kx=ma
What did you take for a ?
Or, equivalently: what for x ?
 
  • #3
Izwx2000 said:
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
BvU said:
Hello Izwx, :welcome:
What did you take for a ?
Or, equivalently: what for x ?
I got 15.38 ms^-2
 
  • #5
And is that valid for the whole launching interval ?
 
  • #6
haruspex said:
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
BvU said:
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
 
  • #8
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
BvU said:
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
 

Related to Find the launch speed of a ball in a spring mechanism

1. How does a spring mechanism work?

A spring mechanism works by storing potential energy in the form of elastic potential energy when it is stretched or compressed. This energy is then released when the spring returns to its original shape, causing the object attached to it to move.

2. What factors affect the launch speed of a ball in a spring mechanism?

The launch speed of a ball in a spring mechanism can be affected by the elasticity of the spring, the mass of the ball, the distance the spring is stretched, and any external forces acting on the system.

3. How can the launch speed of a ball in a spring mechanism be calculated?

The launch speed of a ball in a spring mechanism can be calculated using the equation v = √(k/m)(x), where v is the launch speed, k is the spring constant, m is the mass of the ball, and x is the distance the spring is stretched.

4. Can the launch speed of a ball in a spring mechanism be increased?

Yes, the launch speed of a ball in a spring mechanism can be increased by increasing the distance the spring is stretched or by using a spring with a higher spring constant. However, this also depends on the limitations of the system and how much force the spring can handle.

5. How accurate is the calculation of the launch speed of a ball in a spring mechanism?

The calculation of the launch speed of a ball in a spring mechanism is based on theoretical assumptions and may not always be 100% accurate. Factors such as air resistance and friction can affect the actual launch speed of the ball. The calculated launch speed can be used as an estimate, but the actual speed may vary.

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