Question

Two blocks A and B of mass ma and mb respectively are kept in contact on a frictionless table. The experimenter pushes the block A from behind so that the block accelerate . If the block A exerts a force F on the block B what is the force excreted by the experiment on A?
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Answer 1

Answer:

Explanation:

Let the force exerted by the experimenter on A is F'. Then, the acceleration produced in the system, a = F' / (ma + mb)

 

The force on the block B = F = mb x a = mb x F' / (ma + mb)

 

F' = F (ma + mb)/mb = F (1 + ma/mb)

Answer 2

SOLUTION

Let the force exerted by the experiment be F.

Acceleration produced by this force

$= > \frac{</u><u>F</u><u>}{(</u><u>mA</u><u>+ </u><u>mB</u><u>)}$

Let the force exerted by block A on block B be F'.

$= > \frac{</u></strong><strong><u>F</u></strong><strong><u>}{</u></strong><strong><u>mA</u></strong><strong><u>+ </u></strong><strong><u>mB</u></strong><strong><u>} = \frac{</u></strong><strong><u>F'</u></strong><strong><u>}{</u></strong><strong><u>mB</u></strong><strong><u>} \\ \\ = > </u></strong><strong><u>F</u></strong><strong><u>= </u></strong><strong><u>F'</u></strong><strong><u>[</u></strong><strong><u>1 + \frac{</u></strong><strong><u>mA</u></strong><strong><u>}{</u></strong><strong><u>mB</u></strong><strong><u>} </u></strong><strong><u>]</u></strong><strong><u>$

hope it helps ☺️