## Related questions with answers

Two blocks are in contact on a frictionless table. A horizontal force is applied to one block, as shown in earlier Fig. $(a)$ If $m_1=2.3$ kg, $m_2=1.2$ kg, and $F=3.2$ N, find the force of contact between the two blocks. $(b)$ Show that if the same force $F$ is applied to $m_2$ rather than to $m_1$ , the force of contact between the blocks is $2.1$ N, which is not the same value de- rived in $(a)$. Explain.

Solution

Verified$\textbf{Introduction}$

The two blocks have a common acceleration of

$\begin{align*} a = \frac{F}{(m_1+m_2)}\end{align*}$

where $F$ is the force applied to the first block, $m_1$ is the mass of block 1, and $m_2$ is the mass of block 2.

As force was applied to block 1, a force was also felt by block 2 due to block 1. This force between block 1 and block 2 is the force of contact, denoted as $P$.

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