Refer to the diagrams shown below.
μ = the dynamic coefficient of friction between the block and the level surface.
4mg = the weight of the block.
N = 4mg = the normal reaction force on the block.
R = μN = 4μmg = the frictional force on the block that resists motion.
When the bullet is fired, the total energy of the system is the kinetic energy of the bullet.
The total energy is
E = (1/2)mv²
After the bullet exits the block, it has residual kinetic energy of
KE = (1/2) m*(v/3)² = (1/18) mv²
The remaining energy was used to move the block by a distance d against the frictional force. The work done is
W = R*d
Conservation of energy requires that
E = KE + W
[tex] \frac{mv^{2}}{2} = \frac{mv^{2}}{18} + Rd \\\\ Rd= \frac{8mv^{2}}{18} = \frac{4mv^{2}}{9} \\\\ R= \frac{4mv^{2}}{9d} [/tex]
Answer: The frictional force is [tex] \frac{4mv^{2}}{9d} [/tex]
