Question

A BLOCK OF MASS 2KG IS PLACED ON THE FLOOR. THE COEFFICIENT OF STATIC FRIECTION IS 0.4. A FORCE OF 35N IS APPLIED ON THE BLOCK. CALCULATE THE FORCE OF FRIECTION BETWEEN of and the floor​

Answer 1

Answer:

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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