A 6.0-g bullet leaves the muzzle of a rifle with a speed of 336 m/s. what force (assumed constant) is exerted on the bullet while it is traveling down the 0.83-m-long barrel of the rifle?
The work done on the bullet is force•distance. This is also equivalent to the energy it has leaving the muzzle (work energy theorem). This kinetic energy is (1/2)(0.006kg)(336m/s)^2=338.7J. Then 338.7J=F•0.83m F=338.7/0.83=408N
