Sum

Two blocks A and B of mass m_{A} and m_{B} , 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 = m_{B}a ...(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

Is there an error in this question or solution?

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