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 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.

Explanation:

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Answer 2

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