A6.3 g bullet leaves the muzzle of a rifle with a speed of 596.2 m/s. what constant force is exerted on the bullet while it is traveling down the 0.6 m length of the barrel of the rifle? answer in units of n.

anwser will be 

F = ma

where 

F = force exerted on the bullet 
m = mass of the bullet = 5 gm (given) = 0.005 kg. 
a = acceleration of the bullet 

Substituting appropriately, 

F = 0.005a --- call this Equation 1 

Next working equation is 

Vf^2 - Vo^2 = 2as 

where 

Vf = velocity of the bullet as it leaves the muzzle = 326 m/sec (given) 
Vo = initial velocity of bullet = 0 
a = acceleration of bullet 
s = length of the rifle's barrel 

Substituting appropriately, 

326^2 - 0 = 2(a)(0.83) 

a = 64,022 m/sec^2 

the anwser will be
Substituting this into Equation 1, 

F = 0.005(64,022) 

F =320.11 Newtons 

Hope this helps. 

Choice A is correct.Kinetic energy equation:   KE = (1/2)(m)(v²)This tells us that KE is directly proportional to mass and the square of velocity. In other words, the more mass and more velocity an object has, the more kinetic energy.If an object is sitting at the top of a ramp, there is no velocity and therefore no kinetic energy.    Choices B and D are wrong.A golf ball has more mass than a ping-pong ball, so a ping-pong ball would have less kinetic energy than a golf ball rolling off the end of a ramp.    Choice C is wrong.Choice A is correct.