A 4.4g bullet leaves the muzzle of a rifle with a speed of 322 m/s. What total constant force is exerted on the bullet while it is traveling down the 0.75 m long barrel of the rifle? Answer in...
A 4.4g bullet leaves the muzzle of a rifle with a speed of 322 m/s.
What total constant force is exerted on the bullet while it is traveling down the 0.75 m long barrel of the rifle? Answer in units of N.
To solve this problem, we use the Work-Energy Theorem
Work = change in Kinetic Energy
where Work is
W = F d
and Kinetic Energy is
KE = 1/2 m v^2
The initial KE of the bullet is zero, so the change in KE is found using the mass and final velocity of the bullet. We know the distance over which the force is applied, so the average force on the bullet over that distance is found from
F d = 1/2 m v^2
F = ( 1/2 m v^2 ) / d
F = ( 1/2 ) ( .0044 kg ) ( 322 m/s )^2 / ( .75 m )
F = 304 kg m/s^2
The average force exerted on the bullet is 304 Newtons.
If you have not yet studied the Work-Energy Theorem, the problem can also be solved using kinematics. We can find the acceleration of the bullet using the equation
2 a d = v^2
where a is the acceleration, d the distance traveled, and v the final velocity. Once we have the acceleration, Newton's Second Law gives us the force:
F = m a
where m is the mass of the bullet (in kilograms) and a the acceleration we just found.
The muzzle gained a velocity of 322m/s from its initial velocity of 0, while it travelled through the length s of the barrel which is equal to 0.75m.
So, the initial velocity of the muzzle u=0 and final velocity v= 322m/second and displacement of s=0.75 m is required to gain the velocity of 322m/s. So we use v^2-u^2=2as to find the a, the acceleration.
Therefore, the acceleration of the muzzle in the barrel = (v^2-u^2)/(2s) = (322^2-0)/(2*0.75)= 69122.66666..m/s^2.
The mass of the muzzle = 4.4gram = 0.0044 kg
Therefore the constant force exerted on the muzzle in the barrel = mass of the muzzle*its acceleration in the barrel =0.0044*69122.66666.. N=304.139733..N