How Does Frictional Force Affect Bullet Penetration in Wood?

[mugzieee]

Frictional Force, PLEASE HELP!

A rifle with a barrel length of 55.0cm fires a 12.0g bullet with a horizontal speed of 350m/s. The bullet strikes a block of wood and penetrates to a depth of 11.0cm. What frictional force (assumed to be constant) does the wood exert on the bullet?

heres what i do but its wrong:
F_k_=mu_k_*F_n_

 

[HallsofIvy]

That formula certainly won't work. It's for something moving over a rough surface.

Do this instead: assuming (as the problem says) that the friction force is a constant, F, so that it has acceleration a= F/m, the velocity, t seconds after the bullet hits the wood, is v(t)= v₀- (F/m)t and the distance moved in t seconds is x(t)= v₀t- (F/2m)t².

We know that v<sub>0[/sub= 350 m/s and that the bullet penetrates 11 cm into the wood. Of course, the velocity when it has reached 11 cm is 0.

Solve v(t)= 350- (F/55)t to find the time, t, it takes the bullet to penetrate the wood (depending on F of course). Now put that value of t into 350t- (F/2(12))t²= 11.0 and solve for F.</sub>

 

[xanthym]

A rifle with a barrel length of 55.0cm fires a 12.0g bullet with a horizontal speed of 350m/s. The bullet strikes a block of wood and penetrates to a depth of 11.0cm. What frictional force (assumed to be constant) does the wood exert on the bullet?

heres what i do but its wrong:
F_k_=mu_k_*F_n_

Conservation of Energy can also be used:

{Delta Kinetic Energy} = {Frictional Force Work}

{Delta Kinetic Energy} = (1/2)*m*(v_f)^2 - (1/2)*m*(v_i)^2 =
= 0 - (1/2)*(12x10^(-3) kg)*(350 m/sec)^2
= (-735 Joules)

{Frictional Force Work} = {Friction Force}*(11x10^(-2) meters)

⇒ {Friction Force}*(11x10^(-2)) = (-735 J)
⇒ {Friction Force} = (-6682 N)

~~

 

[mugzieee]

finding F is the hardest part for me. how do i solve for F?>