Question

Three masses are connected via light strings and an ideal pulley, as shown below. Mass m2 is twice as massive as m1; m3 is three times as massive as m1. Everything is initially at rest, but then m3 is released, and it falls 1.30m to the floor. If the speed of m3 just before it hits the floor is 3 m/s, what is the coefficient of kinetic friction between the table and the blocks? (Hint: Use an energy analysis)

Answer 1

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

$\mathbf{-\mu_k ( m_1 + m_2) g h = \dfrac{1}{2} (m_1 + m_2 + m_3)v^2 - m_3gh}$

where;

• m₂ = 2m₁
• m₃  = 3m₁

∴

$\mathbf{-\mu_k ( m_1 + 2m_1)\times g \times 1.3 = \dfrac{1}{2} (m_1 + 2m_1+ 3m_1)3^2 - 3m_1\times 9.8 \times 1.3}$

$\mathbf{-\mu_k ( 3m_1 )\times g \times 1.3 = \dfrac{1}{2} \times 6 m_1 \times 9 - 3m_1\times 9.8 \times 1.3}$

$\mathbf{-\mu_k ( 3m_1 )\times g \times 1.3 =27 m_1 -38.22 m_1}$

$\mathbf{-\mu_k38.22m_1 =27 m_1 -38.22 m_1}$

$\mathbf{-\mu_k38.22m_1 =-11.22m_1}$

$\mathbf{\mu_k =\dfrac{-11.22m_1}{38.22m_1}}$

$\mathbf{\mu_k = 0.29 }$

Therefore, the coefficient of the kinetic friction is 0.29

Learn more about the coefficient of kinetic friction here:

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