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Standard XII

Physics

Second Law of Motion

In the shown ...

Question

In the shown arrangement mass of A = 1 kg, mass of B = 2 kg. Coefficient of friction between A and B = 0.2.

There is no friction between B and ground. The frictional force exerted by A on B equals.

2 N

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3 N

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4 N

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5 N

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Solution

The correct option is A 2 N
Block A moves due to friction. Maximum acceleration of A can be $\frac{f_{m a x}}{m} o r \frac{μ m g}{m} o r μ g = 0.2 \times 10 = 2 m / s^{2} .$ If both the blocks move together, then combined acceleration of A and B can be $\frac{10}{3} o f 3.33 m / s^{2} .$ Since this is more than the maximum acceleration of A, slipping between them will take place and force of friction will be maximum or $μ m_{A} g = 2 N .$

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In the shown arrangement mass of A = 1 kg, mass of B = 2 kg. Coefficient of friction between A and B = 0.2. There is no friction between B and ground. The frictional force exerted by A on B equals.

In the shown arrangement mass of A = 1 kg, mass of B = 2 kg. Coefficient of friction between A and B = 0.2.

There is no friction between B and ground. The frictional force exerted by A on B equals.