The barge of mass 5.0 x 10^6 kg feels a total applied force of 4.5 x 10^5 N and is observed to accelerate at 7.5 x 10^-2 m/s^2. What is the drag force of the water on it?

1. The drag acts opposite to the applied force, so the problem is one-dimensional along that line. 2. The net force is the applied force minus the drag force. 3. Newton's second law sets the net force equal to the mass times the acceleration. 4. Rearranged, the drag equals the applied force minus the mass times the acceleration. 5. Substitute the values: (5.0 x 10^6 kg)(7.5 x 10^-2 m/s^2) = 3.75 x 10^5 N. 6. Subtract: 4.5 x 10^5 N - 3.75 x 10^5 N = 7.5 x 10^4 N. 7. Reporting 3.8 x 10^5 N stops at the mass times acceleration step and never subtracts. _Source: OpenStax College Physics (CC BY 4.0), Ch 4 "Dynamics: Force and Newton's Laws of Motion", section 4.7 Further Applications of Newton's Laws of Motion_