A constant power machine pulls a block on a smooth horizontal...

Power is given by $P = Fv$

Using Newton's second law, $F = m\dfrac{dv}{dt}$

Substituting this into the power equation:

$P = m\left(\dfrac{dv}{dt}\right)v$

$v dv = \dfrac{P}{m} dt$

Integrating both sides, assuming the block starts from rest:

$\int_{0}^{v} v dv = \int_{0}^{t} \dfrac{P}{m} dt$

$\dfrac{v^2}{2} = \dfrac{P}{m} t$

$v = \sqrt{\dfrac{2P}{m}} \sqrt{t}$

Since $P$ and $m$ are constants, $v \propto \sqrt{t}$

Answer: $v \propto \sqrt{t}$