Calculating Kinetic Energy & Average Force of a 1.35g Bullet

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Homework Help Overview

The discussion revolves around calculating the kinetic energy of a bullet and determining the average force exerted on it while moving through a gun barrel. The subject area includes concepts from mechanics, specifically kinetic energy and force calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate kinetic energy using the mass and velocity of the bullet, while also trying to find the average force based on the bullet's travel distance in the barrel. Some participants question the method for calculating average force and suggest considering the work-energy principle.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the relationships between force, energy, and distance. Some guidance has been offered regarding the work-energy principle, but there is no explicit consensus on the approach to take.

Contextual Notes

There is mention of potential assumptions, such as neglecting frictional losses, which may affect the calculations. The original poster expresses uncertainty about the method for calculating average force, indicating a need for further clarification.

Rawpcgamer
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Part 1.) 1.35g bullet leaves the barrel of the gun going 270m/s. What is its kinetic Energy?
-I converted 1.35 g to kg = 0.00135g
-I arrived with an answer of 49.2075 J

Part 2.) If the length of the barrel of the gun described above is 35cm, find the average force exerted on the bullet to move it the length of the barrel.



What I had attempted to do is Mass X Velocity/ 35 cm
-im not sure if that's the way to find avergae force, because other examples I've seen its divided by time.

thank you for looking at this for me
 
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You know the total energy is the kinetic energy right? The work done by a force is just force*distance.
 
im sorry, i understand what your saying but i need an thorough explanation
 
Rawpcgamer said:
im sorry, i understand what your saying but i need an thorough explanation

If you assume no losses due to friction, then by conservation of mechanical energy, shouldn't the work done = change in kinetic energy?
 

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