First, resolve the force into horizontal and vertical components. The horizontal component of the force is given by:

\(F_x = 12 \cos 25^{\circ}\)

Using Newton's second law, \(F = ma\), where \(m = 3 \text{ kg}\), we have:

\(12 \cos 25^{\circ} = 3a\)

Solving for \(a\):

\(a = \frac{12 \cos 25^{\circ}}{3} = 4 \cos 25^{\circ} \approx 3.625 \text{ m/s}^2\)

Using the equation of motion \(s = ut + \frac{1}{2}at^2\), where initial velocity \(u = 0\) and \(t = 5 \text{ s}\):

\(s = \frac{1}{2} \times 4 \cos 25^{\circ} \times 5^2\)

Calculating \(s\):

\(s = \frac{1}{2} \times 4 \times 3.625 \times 25 \approx 45.3 \text{ m}\)