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Question 4

A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. If the minimum force that can be applied on A so that both the blocks move together is 12 N, the maximum force that can be applied on B for the blocks to move together will be:

Solution

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$$f = f_k = \mu_k R = \mu_k m_1\text{ }g$$

$$12 = f_k = \mu_k \times 4\text{ }g$$

$$\therefore \mu_k = \frac{12}{4\text{ }g} = \frac{3}{g}$$

As block $$B$$ is on smooth surface, therefore to move $$A$$ and $$B$$ together, maximum force $$F$$ required to be applied on $$B = $$ frictional force applied on A by B

$$F = \frac{3}{g}(4 + 5)\text{ }g = 27\text{ }N$$

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