A 50g bullet traveling horizontally at 200 m/s embeds itself in a much more massive wooden block initially at rest on a horizontal surface ( [tex]\mu[/tex]k = 0.10). The block then slides 1.2 m toward an ideal spring and collides with it. The block compresses the spring (k=600.0 N/m) a maximum of 20.0 cm. Calculate the mass of the block of wood.
let (b) represent the bullet and (w) the wood block:
[tex]\Delta[/tex]d=1.2 m(+ 0.2 m)?
My thinking is that the initial energy of the bullet does not equal the final energy of the system because the bullet would lose energy when it hits the block (through heat, sound etc).
to find an equation to substitute for Vw+b:
Mw0 + 0.05(200)=(Mw+ 0.05)Vw+b
Vw+b= 10/(Mw + 0.05)
now to put that into the conservation of energy equation:
0.5Mw+b(Vw+b)2= [tex]\mu[/tex]k(Fn)[tex]\Delta[/tex]d + 0.5kx2
can someone tell me if i'm on the right path? i have no clue how to complete the left side algebraically, so it'd be nice to know
if it is even a necessary effort...
also i don't know if the displacement should include the 20 cm the block travels after hitting the spring because at that point it's not just friction acting against the block