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

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The discussion focuses on calculating the force exerted on a bullet as it travels through a rifle barrel. A user presents their calculation using the equation F = (m(v^2)) / (2d) and arrives at a force of 312.2N for a 5.0g bullet traveling at 320m/s down a .82m barrel. Another participant confirms that the calculation is correct and suggests an alternative method using the work-energy principle, equating work done to kinetic energy. The consensus is that the original calculation is accurate, validating the approach taken. The conversation emphasizes the importance of using the correct formula for such physics problems.
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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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