a block of mass of 2 kg is placed on the floor. the coefficient of static friction is 0.4 and there's a force F of 2.5 N applied on the block. calculate the force of friction between block and floor.
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⭕️The applied force is less than the maximum static frictional force.
coefficient of static friction is (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
=0.8×9.8
= fmax = 7.84 N
⭕️The frictional force on the block is equal to the applied force = 2.5 N.
⭕️This static friction is a self adjusting force.
⭕️Static friction is a force that keeps an object at rest.
coefficient of static friction is (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
=0.8×9.8
= fmax = 7.84 N
⭕️The frictional force on the block is equal to the applied force = 2.5 N.
⭕️This static friction is a self adjusting force.
⭕️Static friction is a force that keeps an object at rest.
Answered by
3
Answer:
2.5 N.
Explanation:
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