A bullet of mass 40 g moving with a speed of 90 $m{s}^{-1}$ enters a heavy wooden block and is stopped after a distance of 60 cm. The average resistive force exerted by the block on the bullet is

Here, u = 90 $m{s}^{-1}$,  v = 0

     m = 40 g   $\frac{400}{1000}kg$ = 0.04 kg

s = 60 cm = 0.6 m

Using $v^2-u^2 = 2as$

${\left(0\right)}^2$- ${\left(90\right)}^2$ = 2a $\times$0.6

a = $\frac{{\left(90\right)}^2}{2\times 0.6}$ = -6750 $m{s}^{-2}$

-ve sign shows the retardation.

The average resistive force exerted by block on the bullet is 

F = m $\times$a = (0.04 kg) (6750 $m{s}^{-2}$) = 270 N