Question

3) A block weighting 10 kg rests on a horizontal surface. The coefficient of static
friction between the block and the surface is 0.4 and the coefficient of sliding
friction is 0.2. If a horizontal force of 5N is applied on the block, the magnitude
of the frictional force acting on the block is n N. Then n is​

Answer 1

Answer:

Maximum static frictional force

= 0.4 × 10 × 10 = 40 N (f = umg)(u -> Coefficient of static friction)

Since applied force is less than maximum static frictional force hence-

Frictional force acting on the object

= 5 N (Self- Adjusting )

Thus,

n = 5

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