Force Needed to Stop a Flying Bullet

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SUMMARY

The discussion focuses on calculating the force exerted by a block of wood on a bullet of mass 0.0025 kg, traveling at 470 m/s, which penetrates 6.2 cm before stopping. The relevant kinematic equation used is v² = u² + 2as, where 'v' is the final velocity, 'u' is the initial velocity, 'a' is acceleration, and 's' is the distance traveled. The participants emphasize the importance of converting units to meters and seconds and highlight that the force can be calculated using F = mass * acceleration after determining the acceleration from the kinematic equation.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Familiarity with kinematic equations, specifically v² = u² + 2as
  • Knowledge of unit conversion, particularly from centimeters to meters and from Newtons to kilonewtons
  • Basic principles of motion and deceleration
NEXT STEPS
  • Learn how to apply kinematic equations in real-world scenarios
  • Study unit conversion techniques for physics calculations
  • Explore the relationship between force, mass, and acceleration in various contexts
  • Investigate the effects of different materials on bullet deceleration
USEFUL FOR

Physics students, engineers, and anyone interested in the dynamics of projectile motion and impact forces.

IBdoomed
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A bullet of mass 0.0025 kg moving at 470 m/s
impacts a large fixed block of wood and travels
6.2 cm before coming to rest.
Assuming that the deceleration of the bullet
is constant, find the force exerted by the wood
on the bullet.
Answer in units of kN

I know force= mass * acceleration but i don't know how the distance plays into the solution. Any kind of kinematics formula would be sufficiently helpful!
 
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IBdoomed said:
A bullet of mass 0.0025 kg moving at 470 m/s
impacts a large fixed block of wood and travels
6.2 cm before coming to rest.
Assuming that the deceleration of the bullet
is constant, find the force exerted by the wood
on the bullet.
Answer in units of kN

I know force= mass * acceleration but i don't know how the distance plays into the solution. Any kind of kinematics formula would be sufficiently helpful!

Choose wisely from the sets of equations shown in the first two sections of this wikipedia posting.

http://en.wikipedia.org/wiki/Equations_of_motion
 
i was trying to use v^2= u^2+2as
but the answer i am getting is wrong...
is there a different formula you suggest?
 
IBdoomed said:
i was trying to use v^2= u^2+2as
but the answer i am getting is wrong...
is there a different formula you suggest?

That is certainly the one I would use.

Were you converting everything to metres and seconds?

This formula will only calculate the acceleration of course - you need at least one more step to get the force.
 
nevermind. i was not converting my answer into kN... thank you so much!
 

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