Find the force exerted on a bullet as it travels through the barrel

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SUMMARY

The discussion focuses on calculating the constant force exerted on a 5.1 g bullet traveling at a speed of 467 m/s through a 0.5 m barrel. Using the kinematic equation V² = Vo² + 2a(x - xo), the acceleration (a) is determined to be 218089 m/s². Subsequently, applying Newton's second law (F = ma), the force is calculated to be approximately 1112.25 N. This analysis provides a clear method for determining the force based on the bullet's mass and acceleration.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Familiarity with kinematic equations
  • Basic knowledge of unit conversions (grams to kilograms)
  • Ability to perform algebraic manipulations
NEXT STEPS
  • Study advanced kinematic equations for varying acceleration scenarios
  • Explore the implications of barrel length on bullet velocity
  • Learn about the effects of friction and air resistance on projectile motion
  • Investigate the relationship between force, mass, and acceleration in different contexts
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in ballistics or the dynamics of projectile motion will benefit from this discussion.

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Homework Statement



A 5.1 g bullet leaves the muzzle of a rifle with
a speed of 467 m/s.
What constant force is exerted on the bullet
while it is traveling down the 0.5 m length of
the barrel of the rifle?
Answer in units of N

Homework Equations


V2=Vo2+2a(x-xo)
F=ma

The Attempt at a Solution


4672=2a(.5-0)
a=218089

F=.051*218089
F= 1112.2539
 
Last edited:
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Looks right.
 

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