A cruise ship of mass 10^5 kg sits at rest in port. The captain starts the propellors, which exert a force F = 10^4 Newtons for a distance of 1 km. At the end of thæt distance, the ship has kinetic energy KE = 6.2 x 10^6 Joules.
How much work did the water do on the ship?
Answer: The work done by the propellors on the ship is W(prop) = F * d = (10^4 N) * (1000 m) = 10^7 Joules The total kinetic energy of the ship is smaller than this, though, because the water has been doing negative work on the ship: change in KE = W(prop) + W(water) +6.2 x 10^6 J = 10^7 J + W(water) --> W(water) = 6.2 x 10^6 J - 10^7 J = -3.8 x 10^6 J
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