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A 2 kg block (A) is placed on 8 kg of block (B) which rests on a table. Coefficient of friction between (A) and (B) is 0.2 and between (B) and table is 0.5.A 25N horizontal force is applied on the block (B), then the friction force between the blocks (A) and (B) is :-

A Zero
B 3.9 N
C 5N
D 49 N

User Louis XIV
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Final answer:

The friction force between the 2 kg block A and the 8 kg block B, when a horizontal force is applied to block B, is 3.92 N, corresponding to option B, 3.9 N.

Step-by-step explanation:

To determine the friction force between block A (2 kg) and block B (8 kg) when a horizontal force is applied to block B, we need to consider the coefficient of friction and the normal force. Since the blocks remain together and the coefficient of friction between A and B is 0.2, we use the formula f = μN, where μ is the coefficient of friction and N is the normal force. The normal force on A is equal to its weight, which is mg, or 2 kg × 9.8 m/s² = 19.6 N. The maximum frictional force that can be exerted without sliding is thus 0.2 × 19.6 N = 3.92 N. Therefore, the friction force between the blocks A and B is 3.92 N, which is closest to the provided option B, 3.9 N.

User Farrukh Arshad
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