Answer
a) $6.667\times10^3\;\rm N$
b) $6\times10^{-4} \;\rm s$
c) See the figure below.
Work Step by Step
a) We can assume that the air resistance between the refile's barrel and the wooden block is negligible, so the bullet's horizontal velocity component will not change until it hits the wooden block.
The final velocity of the bullet inside the block is zero. The resistive force of wood is constant so the bullet's acceleration is also constant.
We know that the force exerted by the wooden block is given by
$$F_{wood}=m_{bullet}a_{x,bullet}\tag 1$$
Now we need to find the bullet's acceleration.
$$\overbrace{v_{fx}^2}^{0}=v_{ix}^2+2a_x\Delta x$$
$$-v_{ix}^2=2a_x\Delta x$$
Thus,
$$a_x=\dfrac{-v_{ix}^2}{2\Delta x} \tag 2$$
Plugging into (1);
$$F_{wood}=m_{bullet}\cdot \dfrac{-v_{ix}^2}{2\Delta x}$$
Plugging the known;
$$F_{wood}=10\times10^{-3}\cdot \dfrac{-400^2}{2\cdot 0.12}=-\color{red}{\bf 6.667\times10^3}\;\rm N$$
The negative sign is due to the direction of the force.
b)
We can find the time it takes to come to rest by using the kinematic formula of velocity;
$$\overbrace{v_{fx}}^{0} =v_{ix} +a_x\Delta t$$
Thus,
$$\Delta t=\dfrac{-v_{ix} }{a_x}$$
Plugging from (2);
$$\Delta t=\dfrac{-v_{ix} }{\dfrac{-v_{ix}^2}{2\Delta x} }= \dfrac{2\Delta x} { v_{ix} }$$
Plugging the known;
$$\Delta t=\dfrac{2\cdot 0.12} {400}=\color{red}{\bf 6\times10^{-4}}\;\rm s$$
c) The velocity versus time graph of the bullet is given by the formula of
$$v_f=v_i-a_xt$$
$$\boxed{v=400-666666.67t}$$
