Decelerating force (Physics)

A 1.5g bullet with a speed of 400 m/s strikes a block of wood and comes to rest at a depth of 5 cm.
(a) How large is the average decelerating force?
(b) How long does it take to stop the bullet?

Attempt on a:
mad = (1/2)mv^2
(1.5a)*(0.05) = (1/2)(1.5)(400)^2
a0.075 = 120,000
a = 1.6 x 10^6

You were asked for force and time and instead solved for acceleration.

That aside, your math looks correct. Now just calculate F = ma and t = v/a.
(and be careful of units)

(a) velocity^2 = 2 x a x distance...400^2 = 2 x a x (5/100)....a = 1.6 x 10^6m/sec^2
force = mass x a.....................do the algebra.
(b) velocity = a x time..............time = 400/1.6 x 10^6 seconds...solve

Right track, just finish it:

a = 1.6 x 10^6 m/s^2

(a) F = m a = 0.0015kg * 1.6 x 10^6 m/s^2 = 2.4 x 10^3 N
(b) v = v0 - a t = 0
=> t = v0/a
t = 400 m/s / (1.6x10^6m/s^2) = 0.25 ms