Conservation of momentum with friction

[Thepoint]

Homework Statement

A bullet of mass 1.9×10^−3 kg embeds itself in a wooden block with mass 0.991kg
which then compresses a spring (k= 100N/m ) by a distance 4.5×10^−2m before coming to rest. The coefficient of kinetic friction between the block and table is 0.55.

a) What is the initial speed of the bullet?

b) What fraction of the bullet's initial kinetic energy is dissipated (in damage to the wooden block, rising temperature, etc.) in the collision between the bullet and the block?

Homework Equations

.5kx^2,
normal force x Mu,
.5mv^2,
mgy
m1*v1=(m1+m2)*v2

The Attempt at a Solution

.5kx^2= .5mv^2- (Mu x F(g))
2.25J= .5(.0019kg)v^2 - (.55)1.036kg(9.8m/s^2)
v=91 m/s

I don't know what I did wrong. Also I can answer the second part if I know the first part so just help me out with the first part.

 

[LowlyPion]

Homework Statement

A bullet of mass 1.9×10^−3 kg embeds itself in a wooden block with mass 0.991kg
which then compresses a spring (k= 100N/m ) by a distance 4.5×10^−2m before coming to rest. The coefficient of kinetic friction between the block and table is 0.55.

a) What is the initial speed of the bullet?

b) What fraction of the bullet's initial kinetic energy is dissipated (in damage to the wooden block, rising temperature, etc.) in the collision between the bullet and the block? ]2. Homework Equations
.5kx^2,
normal force x Mu,
.5mv^2,
mgy
m1*v1=(m1+m2)*v2

3. The Attempt at a Solution
.5kx^2= .5mv^2- (Mu x F(g))
2.25J= .5(.0019kg)v^2 - (.55)1.036kg(9.8m/s^2)
v=91 m/s

I don't know what I did wrong. Also I can answer the second part if I know the first part so just help me out with the first part.

Calculate the PE put in the spring:
1/2*100(.045)² = .10 not the 2.25 J you used. (You apparently used 1/2*k*x not x² )

Also m1 + m2 is .9929. Where did the 1.036kg come from?