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Sagot :
The coefficient of the kinetic friction is 0.29
The coefficient of kinetic friction refers to the ratio of the frictional force restricting and opposing the motion of two contact surfaces to the normal force pushing the two surfaces around each other.
From the conservation of energy theorem;
The work done by friction = K.E + P.E
[tex]\mathbf{-\mu_k ( m_1 + m_2) g h = \dfrac{1}{2} (m_1 + m_2 + m_3)v^2 - m_3gh}[/tex]
where;
- m₂ = 2m₁
- m₃ = 3m₁
∴
[tex]\mathbf{-\mu_k ( m_1 + 2m_1)\times g \times 1.3 = \dfrac{1}{2} (m_1 + 2m_1+ 3m_1)3^2 - 3m_1\times 9.8 \times 1.3}[/tex]
[tex]\mathbf{-\mu_k ( 3m_1 )\times g \times 1.3 = \dfrac{1}{2} \times 6 m_1 \times 9 - 3m_1\times 9.8 \times 1.3}[/tex]
[tex]\mathbf{-\mu_k ( 3m_1 )\times g \times 1.3 =27 m_1 -38.22 m_1}[/tex]
[tex]\mathbf{-\mu_k38.22m_1 =27 m_1 -38.22 m_1}[/tex]
[tex]\mathbf{-\mu_k38.22m_1 =-11.22m_1}[/tex]
[tex]\mathbf{\mu_k =\dfrac{-11.22m_1}{38.22m_1}}[/tex]
[tex]\mathbf{\mu_k = 0.29 }[/tex]
Therefore, the coefficient of the kinetic friction is 0.29
Learn more about the coefficient of kinetic friction here:
https://brainly.com/question/13754413?referrer=searchResults
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