Question

A dog, weighing 10.0 lbs, is standing on a flatboat so that he is 20 ft from the shore. He walks 8.0 ft on the boat toward shore and then halts. The boat weighs 40 lb, and one can assume there is no friction between it and the water. How far is he from the shore at the end of this time?

...I proved that the center of mass does not change, which to me, seems like the most entirely unhelpful thing ever. All I know is that there is a difference of 2.0 ft between the initial boat position and the final boat position, but I have no idea whether or not that actually effects where the dog is or not.

I'm really frustrated, because this seems like a simple question, but I just can't conceptually grasp at it.

Answer 1

as you said, the location of the center of mass does not change. So, since the boat weighs 4 times as much as the dog, it moves backward 1 foot for every 4 feet the dog moves forward.

So, as you say, the boat has moved back 2 feet and the dog has moved forward 8 feet.

That means his net motion is 6 ft toward the shore.