Post a New Question


posted by 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?

  • physics - ,

    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

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question