Question

A block of mass 2 kg is kept on the floor. The coefficient of static friction is 0.4. If a force F of 2.5 N is applied on the block as shown in the figure, the frictional force between the block and the floor will be

Answer 1

Answer:

given that ,

》mass of block = 2kg

》force applied = 2.5

》coefficient of static friction, µs = 0.4

☆Maximum static frictional force,

fmax = µs × m × g = 0.4 × 2 × 9.8

=> fmax = 7.84 N

Since,

the applied force is less than the maximum static frictional force,

the frictional force on the block is equal to the applied force = 2.5 N.

☆it is due to the fact that static friction is a self adjusting force.

Answer 2

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Given that ,

》mass of block = 2kg

》force applied = 2.5

》coefficient of static friction, µs = 0.4

☆Maximum static frictional force,

fmax = µs × m × g = 0.4 × 2 × 9.8

=> fmax = 7.84 N

Since,

• the applied force is less than the maximum static frictional force,

the frictional force on the block is equal to the applied force = 2.5 N.

• It is due to the fact that static friction is a self adjusting force.

Thanks..