Draw the free body diagram for block m1.

Explanation:The free body diagram (FBD) for block \( m_1 \) represents all the forces acting on it. These include:Tension (\( T_1 \)): The force exerted by the string pulling upward on \( m_1 \).Weight (\( m_1 g \)): The gravitational force acting downward on \( m_1 \).Acceleration (\( a \)): The block \( m_1 \) accelerates upward, as shown by the arrow labeled \( a \).Force Equation for \( m_1 \):Using Newton’s Second Law (\( F = ma \)):\( T_1 - m_1 g = m_1 a \)\( T_1 = m_1 g + m_1 a \)Free Body Diagram:Conclusion: The free body diagram shows:An upward arrow representing tension (\( T_1 \)).A downward arrow representing weight (\( m_1 g \)).An upward acceleration (\( a \)) of the block.The diagram is correct and accurately represents the forces on \( m_1 \).

Draw the free body diagram for block m1.

Explanation:The free body diagram (FBD) for block \( m_1 \) represents all the forces acting on it. These include:Tension (\( T_1 \)): The force exerted by the string pulling upward on \( m_1 \).Weight (\( m_1 g \)): The gravitational force acting downward on \( m_1 \).Acceleration (\( a \)): The block \( m_1 \) accelerates upward, as shown by the arrow labeled \( a \).Force Equation for \( m_1 \):Using Newton’s Second Law (\( F = ma \)):\( T_1 - m_1 g = m_1 a \)\( T_1 = m_1 g + m_1 a \)Free Body Diagram:Conclusion: The free body diagram shows:An upward arrow representing tension (\( T_1 \)).A downward arrow representing weight (\( m_1 g \)).An upward acceleration (\( a \)) of the block.The diagram is correct and accurately represents the forces on \( m_1 \).

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Draw the free body diagram for block m₁.

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Detailed Solution

Explanation:

The free body diagram (FBD) for block \( m_1 \) represents all the forces acting on it. These include:

Tension (\( T_1 \)): The force exerted by the string pulling upward on \( m_1 \).
Weight (\( m_1 g \)): The gravitational force acting downward on \( m_1 \).
Acceleration (\( a \)): The block \( m_1 \) accelerates upward, as shown by the arrow labeled \( a \).

Force Equation for \( m_1 \):

Using Newton’s Second Law (\( F = ma \)):

\( T_1 - m_1 g = m_1 a \)
\( T_1 = m_1 g + m_1 a \)

Free Body Diagram:

Conclusion: The free body diagram shows: