Why Does Energy Dissipate When a Tug's Cable Breaks?

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When a tug's cable breaks, the tug accelerates while the ship decelerates, illustrating the principle of momentum conservation. The tug's increase in speed occurs because it is no longer exerting force on the ship, allowing it to move freely. Energy is not conserved in this scenario due to the conversion of potential energy in the system into kinetic energy and other forms, such as sound and heat, when the cable snaps. The discussion also touches on the effects of friction and the dynamics of the water, suggesting that these factors play a role in the energy dissipation observed. Understanding these principles is crucial for analyzing the behavior of objects in motion when external forces are suddenly removed.
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Homework Statement


If a tug is pulling a ship and its cable pops it is observed that the tug speeds up while the ship slows down, explain what principle is observed and why energy is NOT conserved.


Homework Equations





The Attempt at a Solution



Principle Of moments and i don't know why energy is NOT conserved
 
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damienr said:

Homework Statement


If a tug is pulling a ship and its cable pops it is observed that the tug speeds up while the ship slows down, explain what principle is observed and why energy is NOT conserved.


Homework Equations





The Attempt at a Solution



Principle Of moments and i don't know why energy is NOT conserved

Don't think moments are involved, but I could be missing something.

Start by thinking of both objects being on a frictionless plane, and you cut the rope...

Then add in friction and think about what happens.

Is water anything like friction?


EDIT -- Welcome to the PF, BTW!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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