# Calculating Kinetic Energy & Force of a 14kg Projectile

• raman911
In summary, the conversation discusses the calculation of the kinetic energy and average force of a projectile fired from a 3 m long cannon with a velocity of 630 m/s and a mass of 14.0 kg. The attempt at a solution includes the correct calculation of the kinetic energy and the average force, but it is noted that the exercise for determining the average force may not have significant physical significance.

## Homework Statement

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.

## 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:
raman911 said:

## Homework Statement

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.

## The Attempt at a Solution

A) Ek=1/2*14kg*630m/s
Ek=2.8*10^6N

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

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.

arildno said:
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.

your mean B is wrong

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

raman911 said:
your mean B is wrong
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.

Saketh said:
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.

thxxxxxxxxxxx