Answer
(a) $\mu_K=0.45$
(b) $E=99\%$
Work Step by Step
(a) The coefficient of kinetic friction can be determined as
First of all, we find the combined velocity of bullet and block
$v=\frac{m_1v_1}{m1_+m_2}$
$\implies v=\frac{0.005\times 200}{0.005+0.75}$
This simplifies to:
$v=1.32m/s$
We also know that $v_f^2=v_i^2+2ad$
As $v_i=0$
$\implies v_f^2=2ad$
$\implies a=\frac{v_f^2}{2d}$
$\implies a=\frac{(1.32)^2}{2(0.2)}=4.36m/s$
As $F=\mu_K mg$
$\implies ma=\mu_K mg$
$\implies \mu_K=\frac{ma}{mg}$
$\implies \mu_K=\frac{a}{g}$
We plug in the known values to obtain:
$\implies =\mu_K=\frac{4.36}{9.81}$
$\mu_K=0.45$
(b) We can calculate the dissipated energy as
$E=1-\frac{K.E_f}{K.E_i}$
$\implies E=1-\frac{m_1}{m_1+m_2}$
We plug in the known values to obtain:
$E=1-\frac{0.005}{0.005+0.75}$
This simplifies to:
$E=0.994$
$E=0.994\times 100 \%=99\%$
