Question

A block of mass 2 kg is placed on the floor. The coefficient of static friction between the two surfaces is 0.4 A. force of 2.5 newton is applied on the block as shown. The force of friction between the block and the floor is

Answer 1

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һєяє' ʏȏȗя ѧṅśwєя ʟȏȏҡıṅɢ ғȏя________
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✯♦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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$hope \: it \: helps$


★BRAINLY★

Answer 2

●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.