(i) Resolve forces down the plane:

\(20 + 5g \, \sin 10^\circ - F = 0\)

\(R = 5g \, \cos 10^\circ\)

\(\mu = \frac{F}{R} = \frac{20 + 8.6824}{49.24}\)

The coefficient of friction \(\mu\) is 0.582.

(ii) Using Newton's second law:

\(5g \, \sin 10^\circ - 0.582 \times 49.24 = 5a\)

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

\(0 = 2.5^2 - 2 \times 4s\)

The distance moved \(s\) is 0.781 m.

Alternative method for part (ii):

Potential energy loss:

\(5g \, d \, \sin 10^\circ\)

\(\frac{1}{2} \times 5 \times 2.5^2 + 5g \, d \, \sin 10^\circ = 0.582 \times 5g \, d \, \cos 10^\circ\)

The distance moved \(d\) is 0.781 m.