In the attached figure, a 4.5 kg dog stands on an 18 kg flatboat at distance D = 6.1 m from the shore. It walks 2.4 m along the boat toward shore and then stops. Assuming no friction between the boat and the water, find how far the dog is then from the shore.

Question

ganeshie8 · Answer

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amistre64 · Answer

i assume the walking pushes the boat thru the water to some degree

ganeshie8 · Answer

yeah dog kicks the boat to right in order to move to left

ganeshie8 · Answer

i have no clue where to start but it seems it has somethign to do with conservation of linear momentum...

ganeshie8 · Answer

answer on back of textbook is 4.2m

amistre64 · Answer

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maybe?

amistre64 · Answer

i dunno how i could make that workable tho

amistre64 · Answer

nothings coming to mind for me, but im not a physics minded person for the most part

empty · Answer

Can we do this:

$$m_1\frac{\Delta x_1}{\Delta t} = m_2\frac{\Delta x_2}{\Delta t}$$ then the time goes away.

amistre64 · Answer

that would give us 4.3 in my original thought

amistre64 · Answer

4.5(2.4)/(18) = .6 for the distance moved by the boat

6.1+.6-2.4

ganeshie8 · Answer

yeah it is giving 4.3
thats very close to the textbook answer!, 4.2

empty · Answer

Yeah, this gives the wrong answer, I am calculating (which I guess is the same thing)

$$6.1 - 4.5\frac{2.4}{18}$$

amistre64 · Answer

the boat moves away from shore, not towards it

amistre64 · Answer

assuming correctness :)

the starting point of the dog is at 6.1 + 4.5(2.4)/18

and the dog has moved 2.4 away from it

6.1 + 4.5(2.4)/18 - 2.4

empty · Answer

We're talking about the dog though

amistre64 · Answer

we know how far the dog moves, my 'd' is the distance the boat moves away from shore as the dog pushes against it

jadedry · Answer

Dog walks 2.4 meters, displaced by 1.9 meters? Therefore boat moved back 0.5 meters?

amistre64 · Answer

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