A) ![v=\sqrt{\frac{2\mu_k (m+M)gd}{m}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7B2%5Cmu_k%20%28m%2BM%29gd%7D%7Bm%7D%7D)
The initial kinetic energy of the bullet is given by
![K_i = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=K_i%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
where
m is the mass of the bullet
v is the initial speed of the bullet
Later, the bullet hit the block and remain embedded with it. The bullet and the block (of mass M) slides over a distance d. The magnitude of the force of friction acting on the bullet+block is
![F=\mu_k (m+M) g](https://tex.z-dn.net/?f=F%3D%5Cmu_k%20%28m%2BM%29%20g)
where
is the coefficient of kinetic friction
g is the acceleration of gravity
The work done by the force of friction is
![W=-Fd = -\mu_k (m+M)g d](https://tex.z-dn.net/?f=W%3D-Fd%20%3D%20-%5Cmu_k%20%28m%2BM%29g%20d)
where d is the displacement of the block+bullet. The negative sign is due to the fact that the direction of the force and of the displacement are opposite.
For the law of conservation of energy, the work done must be equal to the change in kinetic energy. Since the final kinetic energy is zero (the bullet with the block comes at rest), we can write:
![W=K_f - K_i = -K_i](https://tex.z-dn.net/?f=W%3DK_f%20-%20K_i%20%3D%20-K_i)
And so
![-\mu_k (m+M) g d = -\frac{1}{2}mv^2](https://tex.z-dn.net/?f=-%5Cmu_k%20%28m%2BM%29%20g%20d%20%3D%20-%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
Solving for v, we find an expression for the bullet speed:
![-\mu_k (m+M) g d = -\frac{1}{2}mv^2\\v=\sqrt{\frac{2\mu_k (m+M)gd}{m}}](https://tex.z-dn.net/?f=-%5Cmu_k%20%28m%2BM%29%20g%20d%20%3D%20-%5Cfrac%7B1%7D%7B2%7Dmv%5E2%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B2%5Cmu_k%20%28m%2BM%29gd%7D%7Bm%7D%7D)
B) 16.8 m/s
As we calculated in the previous part, the speed of the bullet is given by
![v=\sqrt{\frac{2\mu_k (m+M)gd}{m}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7B2%5Cmu_k%20%28m%2BM%29gd%7D%7Bm%7D%7D)
In this part of the problem we have:
m = 9.0 g = 0.009 kg is the mass of the bullet
M = 12 kg is the mass of the block
d = 5.4 cm = 0.054 m is the distance covered by the block+bullet
is the coefficient of friction
g = 9.8 m/s^2 is the acceleration of gravity
Substituting, we find
![v=\sqrt{\frac{2 (0.20) (0.009 kg+12 kg)(9.8 m/s^2)(0.054 m)}{0.009 kg}}=16.8 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7B2%20%280.20%29%20%280.009%20kg%2B12%20kg%29%289.8%20m%2Fs%5E2%29%280.054%20m%29%7D%7B0.009%20kg%7D%7D%3D16.8%20m%2Fs)