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olchik [2.2K]
2 years ago
5

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)

Physics
1 answer:
Elena L [17]2 years ago
8 0

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:

brainly.com/question/13754413?referrer=searchResults

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V = 13m/s

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Explanation:

This question deals with the idea of momentum.

Since the directions of a compass are fixed, we can take south as horizontal and west as vertical. Since both pieces of coconut are of equal mass, then the resultant of the two pieces would be in between South and West and the third piece would be opposite this direction and be in the North-East (i.e. 45 degree from North) direction

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m_{x} = Mass of piece in West direction = m

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mV_{ry}  = m_{y} V_{y} \\ 2mV_{ry}  = m18\\ V_{ry} = 9

mV_{rx}  = m_{x} V_{x} \\ 2mV_{rx}  = m18\\ V_{rx} = 9

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V = 13m/s

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<span>We can answer this using the rotational version of the kinematic equations:</span><span>
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Where:

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ω₀ = initial angular speed

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Substituting the values into equation 1:<span>
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ω₀ = 28.406 rad/s </span><span>


Using equation 2:
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ω = 8.65 rad/s 


</span>

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