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A 10 kg lemur swings on a vine from a point which is 40.0 m above the jungle floor to a point which is 15.0 m above the floor. If the lemur was moving 2 m/s initially, what will its velocity be at the 15 m point?

Respuesta :

MathPhys

Answer:

22.2 m/s

Explanation:

Energy is conserved.

PE + KE = PE + KE

mgh₁ + ½ mv₁² = mgh₂ + ½ mv₂²

gh₁ + ½ v₁² = gh₂ + ½ v₂²

(9.8) (40.0) + ½ (2)² = (9.8) (15.0) + ½ v₂²

v₂ = 22.2 m/s

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