A 2kg block is placed over a 4kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. The acceleration of the two blocks if a horizontal force of 12N is applied to the lower block is g=10ms-2

# A 2kg block is placed over a 4kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. The acceleration of the two blocks if a horizontal force of 12N is applied to the lower block is $\left(\mathrm{g}=10{\mathrm{ms}}^{-2}\right)$

1. A

2. B

3. C

4. D

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### Solution:

The maximum frictional force between the blocks

$\mathrm{f}=0.2\left(20\right)=4\mathrm{N}$

So, the lower block will move with an acceleration 1ms-2.

for upper block.

$\mathrm{a}=\frac{\mathrm{F}-\mathrm{f}}{\mathrm{m}}=\frac{12-4}{2}=4{\mathrm{ms}}^{-2}$

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