(a) The potential energy lost by the child from X to Y is given by:

\(PE_{\text{lost}} = mgh = 25 \times 9.8 \times 1.8 = 450 \text{ J}\)

The work-energy principle states that the change in kinetic energy is equal to the work done against resistance plus the potential energy lost:

\(\frac{1}{2} mv^2 = PE_{\text{lost}} - \text{Work done}\)

\(\frac{1}{2} \times 25 \times v^2 = 450 - 50\)

\(\frac{1}{2} \times 25 \times v^2 = 400\)

\(v^2 = \frac{400 \times 2}{25} = 32\)

\(v = \sqrt{32} = 4 \text{ m/s}\)

(b) Using energy methods, the potential energy gained from Y to Z is:

\(PE_{\text{gained}} = 25 \times g \times 2 \times 0.28 = 140 \text{ J}\)

The kinetic energy at Y is:

\(KE = \frac{1}{2} \times 25 \times (4)^2 = 200 \text{ J}\)

The work-energy principle for Y to Z gives:

\(F \times 2 = 25 \times g \times 2 \times 0.28 + 200\)

\(F \times 2 = 270\)

The normal reaction \(R\) is:

\(R = 25 \times g \times \cos \alpha = 25 \times 9.8 \times 0.96 = 240 \text{ N}\)

The frictional force \(F = \mu R\):

\(\mu = \frac{F}{R} = \frac{270}{240} = \frac{9}{8}\)