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Question

In an Atwood machine m1 = 6 kg, m2 = 5 kg. Acceleration is (g = 10 m/s²):

RankGuru · Accepted Answer

For an Atwood machine consisting of two masses $m_1$ and $m_2$ connected by a light string over a frictionless pulley ($m_1 > m_2$), the acceleration $a$ is:
$a = \frac{(m_1 - m_2)g}{m_1 + m_2}$
1. **Numerator**: The net driving force is the difference in weights: $m_1 g - m_2 g = (6-5)g = 1g$.
2. **Denominator**: The total mass being accelerated is $m_1 + m_2 = 6 + 5 = 11 	ext{ kg}$.
Substituting:
$a = \frac{1 g}{11} = \frac{1}{11} g$.
In terms of value, $a \approx 0.91 	ext{ m/s}^2$.

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