## The change in kinetic energy of the cabinet is equal to the net work done on the cabinet, which is equal to the force of kinetic friction times the distance over which the force acts:

$\Delta KE = F_{k}d = (87\text{ N})(1.2\text{ m}) = 104.4\text{ J}$

The initial kinetic energy of the cabinet is zero since it starts from rest, so the final kinetic energy of the cabinet is also 104.4 J. The kinetic energy of the cabinet is given by

$KE = \frac{1}{2}mv^{2}$

where $m$ is the mass of the cabinet and $v$ is its speed. Solving for $v$, we find

$v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2(104.4\text{ J})}{22.0\text{ kg}}} = \boxed{3.0\text{ m/s}}$