Physics Homework about force...someone help?

Question

A 4.3 g bullet leaves the muzzle of a rifle with a speed of 304 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.86 m long barrel of the rifle?

Answer 1

we know m=0.0043kg, Vfinal=304m/s, d=0.86m and Vinitial=0

what we need is the value of acceleration ( remember F=ma )

so the formula we need is

vfina^2=viinitial^2 + 2ad
304^2=0 + 2*0.86*a
a=(304^2)/(2*0.86)
a=53730.23256

so F=m*a=53730.23256*.0043=231.04 N

Force is 231.04 N

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Here's an alterative if you want to double check

the kinetic energy (KE)=1/2 *m*(velocity)^2

the KE of the bullet must equal the Work needed to fire it

KE=0.5*0.0043*304^2 =198.6944 j

Work=F*d=198.6944

so F=(198.6944)/0.86=231.04 N

Again we get force = 231.04 N

So we're pretty sure this is the correct answer

Answer 2

I feel nostalgic answering this question. This is falls in the realm of subject called "Internal Ballistics" which I studied long long ago. The solution comes from what are called "Hunt and Hinds" equations involving lot of differential equations and is quite involved. Let us however solve this in a very simple way using kinematic equations.
v^2-u^2=2as
v=304m/sec
u=zero
s=.86m
a=?
acceleration
=v^2/2s
=304^2/2*.86 m/sec^2
force
=mass*acceleration
=[4.3/1000kg]*304^2/2*.86m/sec^2
=462.08 Newtons

Answer 3

vf^2=vi^2 + 2ax and solve for a. Use 304m/s as vf, and 0 as vi. Use 0.86m as the distance x. Then, F=ma, and m=4.3x10^-3 kg, because the unit of force is kgm/s^2 aka a Newton (N).