Calculating the Velocity of a Catapulted Object

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SUMMARY

The discussion focuses on calculating the maximum speed of a stone launched by a catapult using the principles of physics. The catapult is pulled back 10 cm with a force of 30 N, and the stone has a mass of 0.5 kg. The kinetic energy (Ek) is calculated using the formula Ek = 1/2 mv², leading to the conclusion that work done (WD) is equal to kinetic energy, thus WD = 300 J. The conservation of energy principle is applied to equate WD and Ek, allowing for the determination of the stone's velocity.

PREREQUISITES
  • Understanding of kinetic energy formula (Ek = 1/2 mv²)
  • Knowledge of work done calculation (WD = Fs)
  • Familiarity with mass, force, and displacement units
  • Basic principles of conservation of energy
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  • Learn how to apply the conservation of energy in projectile motion
  • Study the relationship between work done and kinetic energy in physics
  • Explore the effects of air resistance on projectile motion
  • Investigate advanced projectile motion equations and their applications
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Students studying physics, particularly those preparing for GCSE exams, as well as educators looking for practical examples of energy conservation and projectile motion calculations.

I-Love-Maths2
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A catapult is pulled back 10cm with a force of 30N. How fast does the stone of mass 0.5kg got at maximum speed assuming no air resistance.

I have absolutely no idea how to do this...
But, I am assuming you use Ek=1/2mv^2, though I can't figure out how to apply it

To rearrange that:
V=sqrt[Ek/(1/2m)]

So, to make it clearer
- 10cm would be the displacement (s)
- 30N would be the force (F)
- 0.5kg would be the mass (m)
- 9.8m/s is the given unit for gravity in GCSE (g)

Other equations that might be useful:
- WD=Fs so WD=300J
- Ep=mgh
- W=mg

So from that we need an equation that uses WD and m

Can anybody help?
 
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How much force will be applied to the stone - and over what distance?
I think we are supposed to assume that the parts of the catapult have zero mass...
and that the catapult is immovable.
 
I-Love-Maths2 said:
WD=Fs so WD=300J
what becomes of WD once the projectile is away ?

By the way, you do have an error in there ...
 
Keep this down to GCSE standard please
Does WD=Ek according to the conservation of energy?
 
I think the exercise allows you to set the kinetic energy equal to the work done.
I-Love-Maths2 said:
Keep this down to GCSE standard please
I'm not from a different planet, but I'm not familiar with this standard. All I know is you don't get Joules if you multiply N with cm :rolleyes:
 
Oh, and a belated: hello Love, :welcome:
 

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