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Sagot :
Answer:
v = 2.82 m/s
Explanation:
For this exercise we can use the conservation of energy relations.
We place our reference system at the point where block 1 of m₁ = 4 kg
starting point. With the spring compressed
Em₀ = K_e + U₂ = ½ k x² + m₂ g y₂
final point. When block 1 has descended y = - 0.400 m
Em_f = K₂ + U₂ + U₁ = ½ m₂ v² + m₂ g y₂ + m₁ g y
as there is no friction, the energy is conserved
Em₀ = Em_f
½ k x² + m₂ g y₂ = ½ m₂ v² + m₂ g y₂ + m₁ g y
½ k x² - m₁ g y = ½ m₂ v²
v² = [tex]\frac{k}{m_2} x^2 - 2 \frac{m_1}{m_2} \ g y[/tex]
let's calculate
v² = [tex]\frac{180}{6.00} \ 0.300^2 - 2 \ \frac{4.00}{6.00} \ 9.8 \ (- 0.400)[/tex]
v² = 2.7 + 5.23
v = √7.927
v = 2,815 m / s
using of significant figures
v = 2.82 m/s
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