91Ó°ÊÓ

The initial velocity of bullet is $u_b=400\text{″¾}/\text{s}$

The initial velocity of block is $u_w=0$

The distance moved by block is $d=72\text{ c³¾}=0.72\text{″¾}$

The final velocity of bullet is $v_b=190\text{″¾}/\text{s}$

The mass of blockis: $M=0.800\text{ k²µ}$

The mass of bullet is: $m=4\text{ g}=0.004\text{ k²µ}$

The magnitude and direction of velocity of stone after collision is found by using momentum conservation along horizontal and vertical direction.

Apply the momentum conservation to find final velocity of block.

$m{u}_b+M{u}_w=m{v}_b+M{v}_w$

Substitute all the values in the above equation.

$\begin{array}{c}\left(0.004\text{ k²µ}\right)(400\text{″¾}/\text{s})+\left(0.8\text{ k²µ}\right)\left(0\right)=\left(0.004\text{ k²µ}\right)(190\text{″¾}/\text{s})+\left(0.8\text{ k²µ}\right)v_w\\ v_w=1.05\text{″¾}/\text{s}\end{array}$

The acceleration of the block is calculated as:

$v_w^2=u_w^2+2ad$

Here, $a$ is the acceleration of wooden block.

$\begin{array}{c}{(1.05\text{″¾}/\text{s})}^2={\left(0\right)}^2+2a\left(0.72\text{″¾}\right)\\ a=0.77\text{″¾}/{\text{s}}^2\end{array}$

The coefficient of kinetic friction between block and surface isgiven as:

$\mathrm{μ}=\frac{a}{g}$

Here, $g$ is the gravitational acceleration and its value is $9.8\text{″¾}/{\text{s}}^2$.

Substitute all the values in the above equation.

$\begin{array}{l}\mathrm{μ}=\frac{0.77\text{″¾}/{\text{s}}^2}{9.8\text{″¾}/{\text{s}}^2}\\ \mathrm{μ}=0.078\end{array}$

Therefore, the coefficient of kinetic friction between block and surface is $0.078$.

The decrease in kinetic energy of bullet is given as:

$\mathrm{Δ}K=\frac{1}{2}m(v_b^2−u_b^2)$

Substitute all the values in the above equation.

$\begin{array}{l}\mathrm{Δ}K=\frac{1}{2}\left(0.004\text{ k²µ}\right)[{(190\text{″¾}/\text{s})}^2−{(400\text{″¾}/\text{s})}^2]\\ \mathrm{Δ}K=−247.8\text{ J}\end{array}$

Therefore, the decrease in kinetic energy of bullet is$−247.8\text{ J}$ .

The kinetic energy of the bullet is given as:

$K=\frac{1}{2}M{v}_w^2$

Substitute all the values in the above equation.

$\begin{array}{l}K=\frac{1}{2}\left(0.8\text{ k²µ}\right){(1.05\text{″¾}/\text{s})}^2\\ K=0.441\text{ J}\end{array}$

Therefore, the kinetic energy of the bullet is $0.441\text{ J}$.