Question

126. A block of mass 4 kg is placed on the floor. The
coefficient of static friction is 0.4. Force of 8 N is
applied on the block. The force of friction between
the block and the floor is
(1) 16 N
(2) 8.ON
(3) 2.0 N
(4) zero​

Answer 1

Answer:

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Answer 2

Answer:-

• Option 1 (16N) is the correct answer.

Solution:-

Given:-

• (i) Mass of block (m) = 4kg
• (ii) Coefficient of static friction (u ) = 0.4
• (iii) External force (F) = 8N
• (iv) g = 10m/s²

Let:-

Maximum force be f.

Then:-

=> f = uN

=> f = u×mg

=> f = 0.4×4×910

=> f = 16 N

Here in this case:-

f > F (body will not move)

f = F

f = 16N