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a force f with arrow applied to an object of mass m1 produces an acceleration of 4.00 m/s2. the same force applied to a second object of mass m2 produces an acceleration of 1.40 m/s2.(a) what is the value of the ratio m1/m2?0.35 correct: your answer is correct.(b) if m1 and m2 are combined into one object, find its acceleration under the action of the force f with arrow.1.037 correct: your answer is correct. m/s2

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The value of the ratio m₁/m₂ is 0.35.

The acceleration of the combined object under the action of the force F is approximately 1.037 m/s².

How to find ratio and acceleration?

(a) Use Newton's second law:


\[ F = m \cdot a \]

For the first object with mass m₁ and acceleration a₁:


\[ F = m_1 \cdot 4.00 \]

For the second object with mass m₂ and acceleration a₂:


\[ F = m_2 \cdot 1.40 \]

Since the same force F is applied in both cases, set the equations equal to each other and solve for the ratio
\( (m_1)/(m_2) \):


\[ m_1 \cdot 4.00 = m_2 \cdot 1.40 \]


\[ (m_1)/(m_2) = (1.40)/(4.00) \]


\[ (m_1)/(m_2) = 0.35 \]

(b) When masses m₁ and m₂ are combined, the total mass
\( m_{\text{total}} \) is m₁ + m₂. The same force F applied to the combined mass will produce a new acceleration
\( a_{\text{total}} \).

Using Newton's second law again:


\[ F = m_{\text{total}} \cdot a_{\text{total}} \]


\( F = m_1 \cdot 4.00 \), so:


\[ m_1 \cdot 4.00 = (m_1 + m_2) \cdot a_{\text{total}} \]

Now, using the ratio
\( (m_1)/(m_2) = 0.35 \) to express m₂ in terms of m₁:


\[ m_2 = (m_1)/(0.35) \]

Substitute m₂ back into the equation:


\[ m_1 \cdot 4.00 = \left(m_1 + (m_1)/(0.35)\right) \cdot a_{\text{total}} \]


\[ 4.00 = \left(1 + (1)/(0.35)\right) \cdot a_{\text{total}} \]


\[ 4.00 = \left(1 + (1)/(0.35)\right) \cdot a_{\text{total}} \]


\[ 4.00 = \left(1 + 2.8571\right) \cdot a_{\text{total}} \]


\[ 4.00 = 3.8571 \cdot a_{\text{total}} \]


\[ a_{\text{total}} = (4.00)/(3.8571) \]


\[ a_{\text{total}} \approx 1.037 \text{ m/s}^2 \]

The acceleration of the combined object under the action of the force F is approximately 1.037 m/s².

User Al Pascual
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