A 2 kg block of wood rests on a long table top. A 5 g bullet moving horizontally with a speed of $150 {\mathrm{ms}}^{-1}$ is shot into the block and sticks to it. The block then slides 2.7 m along the table top and comes to a stop. The force of friction between the block and the table is

Velocity of block just after collision,

                 $v=\frac{5\times {10}^{-3}\times 150}{\left(2+5\times {10}^{-3}\right)}$

                                  (from conservation of linear momentum)

                    $= 0.374 {\mathrm{ms}}^{-1}$

Let F be the force of friction, then work done against friction = initial kinetic energy

or         $F\times 2.7=\frac{1}{2}\times 2.005\times (0.374{)}^2\Rightarrow F=0.052 \mathrm{N}$