First, calculate the change in potential energy (PE) as the block moves from A to B:

\(\text{PE change} = 60g \times 17.5 = 10500 \text{ J}\)

Next, calculate the change in kinetic energy (KE):

\(\text{KE change} = \frac{1}{2} \times 60 \times (8.5^2 - 3.5^2) = 1800 \text{ J}\)

The work done (WD) against resistance is:

\(\text{WD against resistance} = 6 \times 250 = 1500 \text{ J}\)

The total work done by the pulling force is the sum of the changes in PE, KE, and the work done against resistance:

\(\text{WD by pulling force} = 10500 + 1800 + 1500 = 13800 \text{ J}\)

Using the formula for work done by the pulling force:

\(\text{WD by pulling force} = 50 \cos \alpha \times 250\)

Equating the two expressions for work done:

\(50 \cos \alpha \times 250 = 13800\)

Solve for \(\cos \alpha\):

\(\cos \alpha = \frac{13800}{12500} = \frac{102}{125}\)

Finally, calculate \(\alpha\):

\(\alpha = \cos^{-1} \left( \frac{102}{125} \right) \approx 35.3^\circ\)