A block of mass $$2$$ 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 as shown in fig. The force of friction between the block and the floor is (g=9$$⋅$$8m/s2)

Question

A block of mass $$2$$ 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 as shown in fig. The force of friction between the block and the floor is (g=9$$⋅$$8m/s2)

Infinity Learn · Accepted Answer

$$fs=$$μ$$sR=$$μsmg $$=4$$×$$2$$×$$9$$⋅$$8=7$$⋅$$84NThe$$ applied force is less than the limiting friction force. Hence under the applied force, the block does not move. So long, the block does not move, the adjustable friction force is always equal to the applied force. Thus the frictional force is $$2.5$$ N.

A block of mass 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 as shown in fig. The force of
friction between the block and the floor is
$(g = 9 \cdot 8 m / s^{2})$

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A block of mass 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 as shown in fig. The force of friction between the block and the floor is (g=9⋅8m/s2)

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A block of mass 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 as shown in fig. The force of
friction between the block and the floor is
$(g = 9 \cdot 8 m / s^{2})$