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A boy sitting on a boat pulls on a rope with a constant force $F$ over a duration of time $t$. The other end of the rope is either tied to a bridge or to another freely floating boat of equal mass. Does the boy do more work in the case of the bridge, or in the case of the other freely floating boat?

Here's my attempt: Work is defined as $\int F dx$. Since $F$ remains constant for both cases, we only need to analyze the difference in displacement. This is where I am confused. In the case of the bridge, only the boy is moving (the bridge is fixed). In the second case, the boy is moving towards the other boat, and the other boat is also moving towards the other boy. The relative speed of approach would seem to be greater in second case (exactly twice), and since $x = \int v dt$, the boy does more work in the second case.

Is my analysis sound? Is there a better way to deduce the result?

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  • $\begingroup$ Tip: make use of the center of mass of the two boats system. $\endgroup$ – Ignacio Vergara Kausel Jul 3 '14 at 15:47
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    $\begingroup$ To close voters: this seems to be fine to be by the homework policy. The problem is stated in the context of a specific example, but the question is purely conceptual. Voting to leave open. $\endgroup$ – Kyle Oman Jul 3 '14 at 17:37
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You are mostly correct. Since you are only asked for a qualitative result, you can simplify what you said:

If the boy pulls on rope with force $F$ for time $t$, it will move a certain distance $d$. If the other end of the rope is fixed, the work done is $F\cdot d$. If the end of the rope is moving towards him, the total length of rope he reels in is greater than $d$ so more work is done.

Or from conservation of energy: the second boat ends up moving as well - work is needed to effect that motion. The only one doing any work in the system is the boy. So if he is moving two boats he must be doing more work.

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