Copyright © Michael Richmond.
This work is licensed under a Creative Commons License.
Question 1:
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
Copyright © Michael Richmond.
This work is licensed under a Creative Commons License.