Here, in this question, the total momentum of the system (block + bullet) is conserved in the inelastic collision. The kinetic energy will be lost due to work done by the friction after the collision.

Given data

The mass of the bullet is\(m = 28\;{\rm{g}}\).

The mass of the block is\(m' = 1.35\;{\rm{kg}}\).

The distance is\(d = 8.5\;{\rm{m}}\).

The coefficient of friction is \(\mu = 0.28\).

The relation of work done can be written as follows:

\(\begin{array}{c}W = \Delta {\rm{KE}}\\{F_{\rm{f}}} \times d = \frac{1}{2}m{{v'}^2}\end{array}\)

Here,\(v'\)is the combined speed of the system, m is the mass of the bullet, and\({F_{\rm{f}}}\)is the frictional force, whose value is\({F_{\rm{f}}} = \mu N\), where N is the normal force.

Plugging the values in the above equation,

\(\begin{array}{c}\mu N \times d = \frac{1}{2}m{{v'}^2}\\\mu \times mgd = \frac{1}{2}m{{v'}^2}\\v' = \sqrt {2\mu gd} \end{array}\)

The relation to calculate muzzle speed can be written as follows:

\(\begin{array}{c}P = P'\\mv = \left( {m + m'} \right)v'\end{array}\)

Here, v is the required speed.

Plugging the values in the above equation,

\(\begin{array}{c}\left( {28\;{\rm{g}} \times \frac{{1\;{\rm{kg}}}}{{1000\;{\rm{g}}}}} \right)v = \left( {\left( {28\;{\rm{g}} \times \frac{{1\;{\rm{kg}}}}{{1000\;{\rm{g}}}}} \right) + \left( {1.35\;{\rm{kg}}} \right)} \right)\left( {\sqrt {2\left( {0.28} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {8.5\;{\rm{m}}} \right)} } \right)\\v = 336.07\;{\rm{m/s}}\end{array}\)

Thus, \(v = 336.07\;{\rm{m/s}}\) is the muzzle speed.