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

Explanation:

Hope it will help you

Follow me guys

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