Question

A block of mass 2kg is placed on the floor. The coefficient of static friction is 0.4. If a force of 2.8 N is applied on the block parallel to floor. The force of friction between the block and floor ( take g=10 m/s^2

Answer 1

static Friction force acting btwn block n floor is given as, f=uN=0.4*2*10=8N

Since static friction is greater than force applied so friction acting to oppose the force applied is equal to the force itself i.e. 2.8N

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