Calculating Kinetic Energy & Force of a 14kg Projectile

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Homework Help Overview

The discussion revolves around a physics problem involving a 14 kg projectile fired from a 3 m long cannon at a velocity of 630 m/s. Participants are tasked with calculating the kinetic energy of the projectile and the average force acting on it while in the barrel.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Some participants attempt to calculate the kinetic energy using the formula Ek=1/2*mv^2, while others express concern about the physical significance of calculating the average force acting on the projectile during its motion through the barrel.

Discussion Status

There is a mix of attempts to solve the problem and critiques of the exercise's relevance. Some participants question the validity of the average force calculation, suggesting it lacks physical significance, while others acknowledge the correctness of the numerical results despite the concerns raised.

Contextual Notes

Participants note that the exercise may not accurately reflect the dynamics involved in the projectile's motion, particularly regarding the forces acting during the initial phase of firing and the nature of the forces in the barrel.

raman911
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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?
 
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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 [itex]\frac{mv}{2}[/itex] (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.

Bad exercise.
 
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.

Bad exercise.

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
 

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