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P 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 - Physics

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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
  Is there an error in this question or solution?

APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 5 Newton's Laws of Motion
Q 7 | Page 79
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