Using Newton's second law, the acceleration down the slope is given by:

\(a = g (\sin 14^{\circ} - 0.02 \cos 14^{\circ})\)

Substituting the values, we get:

\(a = 9.8 (\sin 14^{\circ} - 0.02 \cos 14^{\circ}) \approx 2.225 \text{ m s}^{-2}\)

Using the equation of motion:

\(v^2 = u^2 + 2as\)

where \(u = 8 \text{ m s}^{-1}\), \(a = 2.225 \text{ m s}^{-2}\), and \(s = 50 \text{ m}\).

Substitute the values:

\(v^2 = 8^2 + 2 \times 2.225 \times 50\)

\(v^2 = 64 + 222.5\)

\(v^2 = 286.5\)

\(v = \sqrt{286.5} \approx 16.9 \text{ m s}^{-1}\)

Thus, the speed of the sledge and the man when they reach the bottom of the track is 16.9 m s^-1.