First, calculate the initial kinetic energy (KE) at the bottom of the hill:

\(\text{Initial KE} = \frac{1}{2} \times 75 \times 10^2 = 3750 \text{ J}\)

Next, calculate the final kinetic energy (KE) at the top of the hill:

\(\text{Final KE} = \frac{1}{2} \times 75 \times 5^2 = 937.5 \text{ J}\)

Calculate the potential energy (PE) gained as the cyclist ascends the hill:

\(\text{PE gained} = 75g \times 700 \times \sin 1.5^{\circ} \approx 13743 \text{ J}\)

Calculate the work done by the force \(F\):

\(\text{WD by } F = F \times 700\)

Using the work-energy principle:

\(\text{WD by } F + \text{Initial KE} = \text{Final KE} + \text{PE gain} + 2000\)

Substitute the known values:

\(F \times 700 + 3750 = 937.5 + 13743 + 2000\)

\(F \times 700 = 16480.5 - 3750\)

\(F \times 700 = 12730.5\)

\(F = \frac{12730.5}{700} \approx 18.5 \text{ N}\)