CBSE-XI-Physics
05: Newton's Laws of Motion
- #7Two 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?Ans : 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\text{ ' }-F={m}_{A}a`` ...(1)
For block B,
F = mBa ...(2)
Dividing equation (1) by (2), we get:
`` \frac{F\text{ ' }}{F}-1=\frac{{m}_{A}}{{m}_{B}}``
`` ``
`` \Rightarrow F\text{ ' }=F\left(1+\frac{{m}_{\,\mathrm{\,A \,}}}{{m}_{\,\mathrm{\,B \,}}}\right)``
∴ Force exerted by the experimenter on block A is `` F\left(1+\frac{{m}_{\,\mathrm{\,A \,}}}{{m}_{\,\mathrm{\,B \,}}}\right)``.
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