Answer: F=m2v2/2MS = 1280 kg-wt.
Explanation:
The velocity of the barrel u at the instant of firing is determined from the law of conservation of momentum. It is given by
u = mv/M,
where m and M are the masses of the projectile and the gun barrel, respectively. The entire kinetic energy 1/2(Mu2) acquired by the barrel at the instant of firing will be expended in work done against the braking force F. Hence
1/2(Mu2) = FS.