(a) The work done against friction is calculated using the formula:

\(WD = F \times d\)

where \(F = 40 \text{ N}\) and \(d = 15 \text{ m}\). Thus,

\(WD = 40 \times 15 = 600 \text{ J}\)

(b) The change in gravitational potential energy is given by:

\(\Delta PE = mgh\)

where \(m = 5 \text{ kg}\), \(g = 10 \text{ m/s}^2\), and \(h = 15 \sin 20^{\circ}\). Therefore,

\(\Delta PE = 5 \times 10 \times 15 \times \sin 20^{\circ} \approx 257 \text{ J}\)

(c) The work done by the pulling force includes both the work against friction and the change in potential energy:

\(WD = 40 \times 15 + 5 \times 10 \times 15 \times \sin 20^{\circ} = 857 \text{ J}\)