Sum
Two blocks A and B of mass mA and mB , respectively, are kept in contact on a frictionless table. The experimenter pushes block A from behind, so that the blocks accelerate. If block A exerts force F on block B, what is the force exerted by the experimenter on block A?
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Solution
Let F' = force exerted by the experimenter on block A and F be the force exerted by block A on block B.
Let a be the acceleration produced in the system.
For block A,
\[F' - F = m_A a\] ...(1)
For block B,
F = mBa ...(2)
Dividing equation (1) by (2), we get:
\[\frac{F'}{F} - 1 = \frac{m_A}{m_B}\]
\[\Rightarrow F' = F\left( 1 + \frac{m_A}{m_B} \right)\]
∴ Force exerted by the experimenter on block A is
\[F\left( 1 + \frac{m_A}{m_B} \right)\]
Concept: Newton’s Second Law of Motion
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