# Kinetic Energy

1. Apr 7, 2007

### raman911

1. The problem statement, all variables and given/known data
A 3 m long canon fires a 14.0 kg projectile with a velocity of 630 m/s [F]. Calculate.
A) The kinetic energy of the projectile as it leaves the canon barrel.
B) The average force acting on the projectile in the barrel.

3. The attempt at a solution
A) Ek=1/2*14kg*630m/s
Ek=2.8*10^6j

B)2.8*10^6j=Fnet*3m
Fnet=2.8*10^6N/3m
Fnet=9.3*10^5N
Is that Right?

Last edited: Apr 7, 2007
2. Apr 7, 2007

### Saketh

Kinetic energy is not $\frac{mv}{2}$ (though your final number is correct), and its unit is the Joule, not the Newton. Other than that, your approach seems correct.

3. Apr 7, 2007

### arildno

I can't possibly see the point in exercise B.
Sure, you can average whatever numbers you want, but that doesn't mean anything if the deviations from that average is humungous, as in this case.

The rate of momentum transfer in the initial phase (from the explosion of gun powder) is orders of magnitude higher than the forces acting from the barrel upon the ball during its motion through it.
In addition, those negligible barrel forces are resistive, rather than contributive.

4. Apr 7, 2007

5. Apr 7, 2007

### arildno

No, I mean that the exercise is dumb, not that you have made a mistake.

6. Apr 7, 2007

### Saketh

He means that the question is asking you to determine a quantity which has little physical significance. Even if you calculate the average force acting on the bullet as kinetic energy is added to it, you will be stuck with a value that means nothing.

But this is the fault of the question, and not the fault of your answer.

7. Apr 7, 2007

### raman911

thxxxxxxxxxxx