Calculating Force: The Physics Behind a Rifle's Shot Speed and Barrel Length

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SUMMARY

The discussion centers on calculating the force exerted on a 5.0 g bullet traveling at 320 m/s down a .82 m barrel. The calculated acceleration is 62439 m/s², leading to a force of 312 N when multiplied by the bullet's mass. The discrepancy with the textbook answer of 310 N is attributed to the consideration of significant figures in the final result. The calculations confirm the application of kinematic equations and Newton's second law in determining force.

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atse1900
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A 5.0 g bullet leaves the muzzle of a rifle with a speed of 320 m/s, What force (assumed constant) is exerted on the bullet while it is traveling down the .82 m long barrel of the rifle?

This is what I have so far. : [.

(320)^2/(2*.82)=t

320/t=a

a= 62439

62439*.005
=312

The back of the book is 310. What did I do wrong?
 
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I got what you got.
 
probably because the book gives answers in significant figures?
 

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