Question

A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. Force of 2.8 N is applied on the block. The force of friction between the block and the floor is (g = 9.8 m/s2)
(1) 2.8 N (2) 7.84 N (3) 2.0 N (4) zero


Answer with explanation!!!!!

Answer 1

2.8 N is frictional force because object is at rest. suppose that the force of friction is exerted from the right and now the actual force is acted from the left so same force will continue to be act and the object will be at rest

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Answer 2

Answer:-

Option 1 (2.8N) is the correct answer.

Solution:-

Given:-

• (i) Mass of block (m) = 2kg
• (ii) Coefficient of static friction (u ) = 0.4
• (iii) External force (F) = 2.8N
• (iv) g = 9.8m/s²

Let:-

• Maximum force be f.

Then:-

=> f = uN

=> f = u×mg

=> f = 0.4×2×9.8

=> f = 7.84 N

Here in this case:-

f > F (body will not move)

f = F

f = 2.8N