Given the power of the engine is constant at 15,000 W, we use the formula for power:

\(P = Fv\)

At point A, the force \(F_A = 750\) N, so:

\(15,000 = 750v_A\)

\(v_A = \frac{15,000}{750} = 20 \text{ m/s}\)

At point B, the force \(F_B = 500\) N, so:

\(15,000 = 500v_B\)

\(v_B = \frac{15,000}{500} = 30 \text{ m/s}\)

The increase in kinetic energy is given by:

\(\Delta KE = \frac{1}{2} m (v_B^2 - v_A^2)\)

\(\Delta KE = \frac{1}{2} \times 1000 \times (30^2 - 20^2)\)

\(\Delta KE = 500 \times (900 - 400)\)

\(\Delta KE = 500 \times 500 = 250,000 \text{ J}\)

Thus, the increase in kinetic energy is 250,000 J or 250 kJ.