# A rifle barrel question related with Kinetic Energy Theorem

1. Oct 16, 2015

### youngC0610

1. The problem statement, all variables and given/known data
A 15.0g bullet is accelerated from rest to a speed of 780m/s in a rifle barrel of length of 72.0cm
a)Find the kinetic energy of the bullet as it leaves the barrel
b)Use the work kinetic energy theorem to find the net work that is done on the bullet.
(this problem had part c.d.e... which were meant to be solved with the length info, and I figured so I didn't post here: length of barrel is unnecessary for part a and b.)
2. Relevant equations
my question is more about conceptual part of the problem: I accidently got the part b right from thinking the initial KE is zero, but part a's answer is 4563J (equal to final KE).
I'm confused how if its initial energy is 4563J and final energy is 4573J results the kinetic energy theorem deltaK = Kf- Ki = 4563J.

3. The attempt at a solution
a) "at rest" so I thought kinetic energy is 0J on the object and another external work cased the bullet to move.
b) Thinking the initial kinetic energy affected the net work would be equal to final kinetic energy : 1/2*.015kg*(780m/s)^2. =4563J

2. Oct 16, 2015

### haruspex

Part a asks for the KE as it leaves the barrel. I.e. after it has been fired.

3. Oct 16, 2015

### JBA

The bullets initial Kinetic Energy is 0 and the source of Potential Energy to be converted to the bullet's Kinetic Energy is stored in the explosive powder inside the the bullet's cartridge shell., i.e. KE bullet = PE explosive powder.