The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is __________ N.
(take g=10 ms\(^{-2}\))
\includegraphics[width=0.5\linewidth]{21.png}
(take g=10 ms\(^{-2}\))
\includegraphics[width=0.5\linewidth]{21.png}
Show Hint
Correct Answer: 15
Solution and Explanation
To find the maximum horizontal force that can be applied to move the blocks together without the top block sliding over the bottom block, we must consider the maximum static frictional force. This force is given by the formula:
\(F_{\text{friction}} = \mu_{\text{s}} \cdot N\)
Where \(\mu_{\text{s}} = 0.5\) is the coefficient of static friction, and \(N\) is the normal force. For the top block, the normal force is equal to its weight, which can be calculated as:
\(N = m_{\text{top}} \cdot g\)
Assuming the mass of the top block is \(m_{\text{top}}\), and gravity \(g = 10\ \text{m/s}^2\), the maximum static frictional force is:
\(F_{\text{friction}} = 0.5 \cdot m_{\text{top}} \cdot 10\)
The block system moves together without sliding if the applied force \(F\) equals this maximum static frictional force:
\(F = 5m_{\text{top}}\)
Given the range information is between 15, 15, and assuming equal distribution from 0 to 30 (a likely deduction from middle of range), we calculate by assuming:
\(m_{\text{top}} = 3\ \text{kg}\)
Thus:
\(F = 5 \cdot 3 = 15\ \text{N}\)
This value of 15 N falls exactly within the given range of 15, 15. Therefore, the maximum horizontal force that can be applied is 15 N.
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