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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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Two blocks of masses $m_1$ and $m_2$, with $m_1 \lt m_2$, are placed in contact with each other on a frictionless, horizontal surface, as shown. A constant horizontal force $F$ is applied to $m_1$ as shown. What is the contact force on $m_2$?

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Answer: $\large\frac{Fm_2}{m_1+m_2}$
Lets draw the Free Body Diagrams for the two masses in contact and the resultant forces:
Here R is the normal reaction force between the blocks.
For block-1, $F - R = m_1a$ and for block-2, $R = m_2a$
Adding the two equations, $F - R + R = F = (m_1+ m_2)a$
$\Rightarrow a = \large\frac{F}{m_1+m_2}$
Therefore we can solve for $R = m_2 a = \large\frac{Fm_2}{m_1+m_2}$
answered Aug 20, 2014 by balaji.thirumalai

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