#### Answer

(a) The block of mass $2m$ slides a distance $d$.
(b) The block slides a distance of $4d$.

#### Work Step by Step

(a) The blocks slide to a stop because of the force of kinetic friction. We can find the rate of deceleration $a_1$ for the block of mass $m$.
$F_f = ma_1$
$mg~\mu_k = ma_1$
$a_1 = g~\mu_k$
We can find the rate of deceleration $a_2$ for the block of mass $2m$.
$F_f = (2m)~a_2$
$(2m)~g~\mu_k = (2m)~a_2$
$a_2 = g~\mu_k$
Since the initial velocity is the same and the rate of deceleration is the same, then both blocks will slide the same distance $d$. The block of mass $2m$ slides a distance $d$.
(b) $d_1 = \frac{0-(v_{0x})^2}{2a_1}$
We can find the distance that the block slides when the initial velocity is $2v_{0x}$.
$d_2 = \frac{0-(2v_{0x})^2}{2a_2}$
$d_2 = 4\times \frac{0-(v_{0x})^2}{2a_1}$
$d_2 = 4~d_1$
The block slides a distance of $4d$.