Posted by diagne on .
A man with a mass of 64.1 kg stands up in a 61-kg canoe of length 4.00 m floating on water. He walks from a point 0.75 m from the back of the canoe to a point 0.75 m from the front of the canoe. Assume negligible friction between the canoe and the water. How far does the canoe move?
The canoe will move so that the center of mass of man plus canoe stays in the same place.
The center of mass in the beginning is a distance X from center, in the back portion.
64.1*1.25 + 61*0 = 125.1 X
X = 1.13 m
When the man walks to within 0.75 m of the front of the canoe, he will have moved 2.5 m relative to the canoe. The center of mass will then be 1.13 m to the front of center.
For the center of mass to remain in the same place relative to the water, the cane must move 2.26 m to the rear