Question

Two boys with masses of 40 kg and 60 kg stand on a horizontal frictionless surface holding the ends of a light 10-m long rod. The boys pull themselves together along the rod. When they meet the 40-kg boy will have moved what distance?.

Answer 1

when they meet the 40-kg boy will have moved a distance of 6 m.

• What is friction force:

Friction is the force that resists motion when the surface of one object comes in contact with the surface of another.

• The principle of center mass.

It is the average position of all the parts of the system, weighted according to their masses.

here,

masses of the boys

M=40 kg,

m=60 kg

x1 = 10 m

The displacement of the 40 kg boy is calculated from the principle of center mass.

XM = ( mx1 + Mx2) / ( m + M )

X(40) = (60 x 10 +  40 x 0) / (40 kg + 60 kg)

X(40) = (600)/(100)

X(40) = 6 m

X(60) = (60 x 0 + 40 x 10 m) / (40 kg + 60 kg)

X(60) = (400)/(100)

X(60) = 4 m

Thus,

when they meet the 40-kg boy will have moved a distance of 6 m.

Learn more about center of mass here:

brainly.com/question/8662931

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