Physics bullet in wood; I can't figure it out?

[x86]

Homework Statement

I can't figure this one out. I can't use any of Newtons laws to solve this, kinetic formulas, and w=Em+Ek doesn't seem to be working for energy. I know the answer, but I can't solve it.

kg = kilogram
m/s = meters per second
m = meter

A 0.002kg bullet is traveling at 87 m/s, after passing through a 4 cm block of wood its speed is 12 m/s.

a) How much work is lost due to friction (Answer is -7.4 J but I can't solve it mathematically)
b) Average force of block (Answer is -180 N)

Homework Equations

a) For work lost due to friction I'm guessing they want to know the air resistance, I have no idea how to do this part

b) Average force of block:
wcd = (.5*12^2)-(.5*87^2) = -3 712.5 J

P = w/t but the only problem is no time is supplied. I'm not quite sure what to do

The Attempt at a Solution

a) .5 * 87^2 = .5 * 12^2

 

[S_Happens]

Work isn't lost, it's "done." Energy is lost. Do you know the work-energy theorem (I'm not sure if the w=Em+Ek is supposed to be some form of that)?

 

[tal444]

a) KE_original must equal the work_friction + KE_after
b) You must find the a that the block causes on the bullet first. Use v₂² = v₁² + 2ad to solve for a.

 

[x86]

Work isn't lost, it's "done." Energy is lost. Do you know the work-energy theorem (I'm not sure if the w=Em+Ek is supposed to be some form of that)?

Yes, W = Ek2 - Ek1 = 1/2m(v2^2-v1^2)
1/2 * 0.002 * (87^2-12^2) = 7.4 J. I can't believe I forgot about that. Thank you.

a) KE_original must equal the work_friction + KE_after
b) You must find the a that the block causes on the bullet first. Use v₂² = v₁² + 2ad to solve for a.

I can't solve for a because I don't have the displacement or such, unless it is 4 CM?

87^2 = 12^2 + 2 * a * 0.04

7425/0.08 = a
a = 92 812.5

F = 0.002 * 92 812.5 = 185.62500 N

Thank you for the help, I really appreciate it

 

[tal444]

Nicely done, and glad to be of help.