A man with a mass of 65.3 kg stands up in a 115-kg canoe of length 4.00 m floati
ID: 2142616 • Letter: A
Question
A man with a mass of 65.3 kg stands up in a 115-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?
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and although it isn't required, I'd appreciate an explaination as well.
Explanation / Answer
the above problem is based on the concept of centre of mass of a systemwhich remains stationary if no external force is applied to the system....in above case there is no external force acting on the system hence the centere of mass must be stationary(at the same point after the man has come forward as before.)
consider the figure
a._______________.b __.c
let a (origin) be the starting point of canoe and b be the intial position of man=3.25 meters and c be the end point of cannoe(ac=4m)
the center of mass of system is given by=(115*2+65.3*3.25)/(115+65.3)............1
now as the man move forward on canoe ....consider the new figure
a.-----a'__.b'______________.c
let us assume that the cannoe moved aa'=x meters from origin and the new position of man be b'and a'b'=.75,ab'=x+.75
now applyinthe center of mass eqn we have
center of mass = (115*(2+x) + 65.3(.75+x))/115+65.3......................2
as i said both the position of center of mass as calculated from eqn 1 and eqn 2 should be same therefore comparing the equations we have
(115*2+65.3*3.25)/(115+65.3)=(115*(2+x) + 65.3(.75+x))/(115+65.3)
we get x= 0.9054 meters
hence the cannoe move 0.9054 meters away frome the origin.....
plzzzzzz rate ....plzzz
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