#### Answer

(a) $F = 6670~N$
(b) $t = 0.600 ~ms$

#### Work Step by Step

(a) We can find the rate of deceleration after the bullet strikes the block.
$a = \frac{0-v_0^2}{2x}$
$a = \frac{-(400~m/s)^2}{(2)(0.12~m)}$
$a = -6.67\times 10^5~m/s^2$
We can use the magnitude of acceleration to find the force.
$F = ma$
$F = (0.010~kg)(6.67\times 10^5~m/s^2)$
$F = 6670~N$
(b) We can find the time it takes the bullet to stop.
$v_f= v_0 + at$
$t = \frac{v_f-v_0}{a}$
$t = \frac{0-400~m/s}{-6.67\times 10^5~m/s^2}$
$t = 0.600~ms$