The discussion centers on calculating the kinetic energy and average force of a 14 kg projectile fired from a 3 m long cannon at a velocity of 630 m/s. The kinetic energy (Ek) is correctly calculated as 2.8 x 10^6 Joules using the formula Ek = 1/2 * m * v^2. However, the average force (Fnet) calculation, resulting in 9.3 x 10^5 Newtons, is critiqued for lacking physical significance due to the nature of forces acting on the projectile during its motion through the barrel.
PREREQUISITESStudents studying physics, particularly those focusing on mechanics, as well as educators seeking to understand common misconceptions in projectile motion calculations.
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.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?
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.
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.raman911 said:your mean B is wrong
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.