a block of mass of 2 kg is placed on the floor. the coefficient of static friction is 0.4 . a force F of 2.5 N is applied on the block. calculate the force of friction between the block and the floor.

We have, coefficient of static friction, µs = 0.4

Mass, m = 2 kg

Force applied, F = 2.5 N

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. This is according to the fact that static friction is a self adjusting force.

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