Is the Calculation of Force on a Bullet in a Rifle Barrel Correct?

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SUMMARY

The calculation of force exerted on a bullet in a rifle barrel is confirmed to be accurate using the formula F = (mv²)/(2d). In this case, a 5.0g bullet traveling at 320m/s in a barrel length of 0.82m results in a force of 312.2N. The discussion highlights two methods for solving the problem, emphasizing the relationship between work done and kinetic energy. The approach using energy principles is validated as correct.

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help on homework please...force

hey, i don't know if what i did for this question is correct, because the equation i found in my textbook may not be the right one to use.

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

i used the equation:
(m(v^2)) / 2d =F

(.005kg*(320m/s)^2 )/ (2*.82m) =F
F= 312.2N

Is this close to being correct?
 
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There are two ways to do this problem, the easiest is with energy (which is what I assume you did), I deleted a post doing the other way :redface:

Take work done on the bullet and equate to kinetic energy:

Fd = \frac{1}{2}mv^2
F = \frac{mv^2}{2d}

Your answer looks good to me :approve:
 

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