1. **State the problem:** A ship initially travels at 18 km/h. After stopping its engines, it drifts 0.6 km and its speed reduces uniformly to 14 km/h. The ship's mass is 2000 t (which is 2,000,000 kg). We are asked to find the force opposing the motion of the ship. 2. **Convert units:** Speeds are given in km/h, convert to m/s for consistent SI units: $$18\text{ km/h} = \frac{18 \times 1000}{3600} = 5 \text{ m/s}$$ $$14\text{ km/h} = \frac{14 \times 1000}{3600} = \frac{14000}{3600} \approx 3.8889 \text{ m/s}$$ Distance drifted: $$0.6 \text{ km} = 600 \text{ m}$$ Mass: $$2000 \text{ t} = 2000 \times 1000 = 2,000,000 \text{ kg}$$ 3. **Calculate acceleration:** Since the speed reduces uniformly, we use kinematics equation: $$v^2 = u^2 + 2as$$ where $u = 5 \text{ m/s}$ (initial speed), $v = 3.8889 \text{ m/s}$ (final speed), $s = 600 \text{ m}$ (distance), $a$ is acceleration (deceleration here). Plug in values: $$3.8889^2 = 5^2 + 2a(600)$$ $$15.123 = 25 + 1200a$$ Solve for $a$: $$1200a = 15.123 - 25 = -9.877$$ $$a = \frac{-9.877}{1200} = -0.00823 \text{ m/s}^2$$ (The negative sign indicates deceleration.) 4. **Calculate opposing force:** Using Newton’s second law, $$F = ma$$ $$F = 2,000,000 \times (-0.00823) = -16460 \text{ N}$$ The negative sign indicates the force direction opposes the motion. **Final answer:** The force opposing the motion of the ship is approximately $$1.65 \times 10^4 \text{ N}$$ acting opposite to the direction of travel.