How Is the Average Force on a Bullet Calculated?

[highc]

Ok, I'm using Schaum's Outline of College Physics to supplement my ultra condensed correspondance physics course. As I work through the "Supplementary Problems" I've come across this one which leaves me puzzled.

Typically a bullet leaves a standard 45 caliber pistol (5.0 inch barrel) at a speed of 262 m/s. If it takes 1 ms to traverse the barrel, determine the average acceleration by the 16.2 g bullet within the gun and then compute the average force exerted on it. The provided answers are: 3.0 x 10^5 m/s, 0.4 x 10 N.

I've had no problem working out the average acceleration to 2.62 x 10^5 m/s (3.0 X 10^5 m/s), but I have no idea how the book has arrived at 40 N for the average force exerted.

Does anybody care to show how this was worked out?

 

[quasar987]

This is obviously an error in the book. With 2.62 x 10^5 m/s² (don't forget to square s) as the average acceleration, the average force, which is given by the proportionality of force and acceleration

$$F_{av}=ma_{av}$$

turns out to be 0.43092 x 10^5 N, which would round down to 0.4 x 10^5 N. So it's probably just a printing error; they forgot the $^5$.

 

[kyle8921]

I would first clarify that the average force exerted on the bullet is not 40 N, but rather 0.4 x 10 N, which is equivalent to 4 N. This is a small but important distinction.

To calculate the average force exerted on the bullet, we can use the equation F=ma, where F is the force, m is the mass, and a is the acceleration. We know the mass of the bullet is 16.2 g, or 0.0162 kg. We also know the average acceleration is 3.0 x 10^5 m/s.

Plugging these values into the equation, we get F = (0.0162 kg)(3.0 x 10^5 m/s^2) = 4.86 N. This is the average force exerted on the bullet as it travels through the barrel of the gun.

It is possible that the book made an error in their calculation or in the units used, resulting in the answer of 40 N. However, with the given values, the correct answer would be 4.86 N.