Question

A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. If a force of 2.8 N is applied on the block parallel to the floor, the force of friction between the block and the floor is (g = 10 m s-2) please help me exject answers.​

Answer 1

Explanation:

Correct option is

A

2.8N

Friction is a self adjusting force. 

the max. value of it here is μmg = 0.4x20= 8 N

But since the applied force is less than 8N. So, the friction adjusts its value and new value becomes 2.8N so that block remains in equllibrium.

Answer 2

Answer:

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