The man of mass m1 = 82 kg and the woman of mass m2 = 80 kg are standing on opposite ends of the platform of mass m0 = 103 kg which moves with negligible friction and is initially at rest with s = 0. the man and woman begin to approach each other. find the displacement s of the platform when the two meet if the displacement x1 of the man relative to the platform is 9 m. the length l of the platform is 15.5 m.

Δs = 0.0679 m

Explanation:

m₀ = 103 kg

m₁ = 82 kg

m₂ = 80 kg

x₀ = (L/2) = (15.5/2) m = 7.75 m

Initial positions

x₁ = 0 m

x₂ = L = 15.5 m

Final positions

x₁ = 9 m

x₂ = L – x₁ = 15.5 m – 9 m = 6.5 m

Xcm = (m₀*X₀+m₁*X₁+m₂*X₂) / (m₀+m₁+m₂)

We get the center of mass in every case

Xcm initial = (103 Kg *7.75 m +82 Kg*0 m+ 80 Kg*15.5 m)/(103+82+80) Kg = 7.6915 m

Xcm final = (103 Kg *7.75 m +82 Kg*9 m+ 80 Kg*6.5 m)/(103+82+80) Kg = 7.7594 m

Variation: Δs = Xcm final - Xcm initial = 7.7594 m - 7.6915 m = 0.0679 m

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explanation:

it was a fall down when it happens

explanation: