#### Answer

$2.1~m$

#### Work Step by Step

We can determine the required distance as follows:
$m_bv_b+m_{sb}v_{sb}=(mb+m_{sb})v$
$\implies (50)(5)+5(0)=(50+5)$
$\implies v=4.545m/s$
According to the law of conservation of energy
$T_A+V_A=T_B+V_B$.eq(1)
$T_A=\frac{1}{2}(m_b+m{sb})v_A^2=\frac{1}{2}(50+5)(4.545)^2=568.068J$
and $V_A=(m_b+m_{sb})gh_A=0$
$T_B=\frac{1}{2}(m_b+m_{sb})v_B^2=0$
$V_B=(m_b+m_{sb})gh_B$
$\implies V_B=(m_b+m_{sb}) gs \space sin 30^{\circ}=55\times 9.81 sin30^{\circ} s=269.775s$
We plug in the known values in eq(1) to obtain:
$568.068+0=0+269.775s$
This simplifies to:
$s=2.1m$