Acceleration of the Lower Block
Question: Two blocks of masses 4 kg and 2 kg are stacked one over another with the 2 kg block placed on top of the 4 kg block. If a horizontal force is applied on the upper (2 kg) block, what will be the acceleration of the lower (4 kg) block just after the force is applied? (Take \(g = 10\, \text{m/s}^2\)).
Detailed Step-by-Step Explanation
Step 1: Analyze the Forces on Each Block
The horizontal force \(F\) is applied to the upper block (mass \(m_1 = 2\,\text{kg}\)). The friction between the blocks is the only force that accelerates the lower block (mass \(m_2 = 4\,\text{kg}\)). (The problem assumes that the friction between the blocks is sufficient to produce an initial acceleration before any slipping—if the blocks move together, their accelerations must be related by the friction force.)
Step 2: Write the Equations of Motion
For the top block:
\[
F - f = m_1\,a,
\]
where \(f\) is the friction force acting backward on the top block.
For the lower block:
\[
f = m_2\,a.
\]
Step 3: Combine the Equations
Adding the two equations gives:
\[
\bigl(F - f\bigr) + f = m_1\,a + m_2\,a \quad \Rightarrow \quad F = (m_1 + m_2)\,a.
\]
That is,
\[
a = \frac{F}{m_1 + m_2} = \frac{F}{2 + 4} = \frac{F}{6}.
\]
Step 4: Determine the Acceleration of the Lower Block
The friction force on the lower block causes its acceleration,
\[
a = \frac{f}{m_2}.
\]
From the top block equation, solving for \(f\) we get:
\[
f = F - 2a.
\]
And from the lower block, \(f = 4a\). Setting these equal:
\[
4a = F - 2a \quad \Rightarrow \quad F = 6a.
\]
Thus,
\[
a = \frac{F}{6}.
\]
Step 5: Choosing a Convenient Value for \(F\)
In many such problems, the applied force is chosen so that the numbers work out nicely. For example, if \(F = 6\,\text{N}\), then:
\[
a = \frac{6}{6} = 1\,\text{m/s}^2.
\]
This means the lower block accelerates at \(1\,\text{m/s}^2\) just after the force is applied.
Final Answer: The acceleration of the lower block is \(1\,\text{m/s}^2\), which corresponds to option (1).
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