Potential and kinetic force of an arrow shot at space.

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SUMMARY

The discussion centers on the physics of an arrow shot in space, where friction and gravitational forces are absent. The force exerted by the bow on the arrow is 2000 N for a duration of 0.05 seconds, resulting in an acceleration of 4000 m/s² and a final velocity of 200 m/s. The conversation also explores the calculation of potential energy (PE) when the arrow is drawn back in the bow and emphasizes the conservation of energy principle, stating that the potential energy before release equals the kinetic energy (KE) after release.

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  • Familiarity with the formula for calculating force (F = m * a)
  • Knowledge of conservation of energy principles
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Karagoz
Assume a person is at space, and assume there's no friction or gravitational force in that space.

The person has a bow and arrow. He stretches the arrow on the bow, and then shots the arrow out in the space. Since there are no friction and gravitational force in that space, the arrow will have constant velocity very long distance or until another force exerted on it.

ArrowAndBow.png
The force exerted on the arrow from the bow is 2000 N. The force is exerted only for 0.05 seconds since the arrow quickly moves out off the bow. And the arrow weights 0.5 Kg. From my calculations* the velocity of the arrow will be 200m/s if I'm not wrong.

But I wonder what will be the potential energy of the arrow when it's pulled in the bow?
How do we calculate it?

*Force = mass * acceleration

1 N = 1 Kg * 1m/s^2

2000 N = 0.5 Kg * a

2000 / 0.5kg = a

a = 4000m/s^2

Velocity = acceleration*time

Velocity = 4000 * 0.05s = 200m/s.

Speed of the arrow in the space is 200m/s.
 

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Karagoz said:
How do we calculate it?

Using conservation of energy. The PE before release equals the KE after release.
 
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