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A block weighing 20 N rests on a horizontal surface. The coeff of static friction b/w the block and surface is 0.04 and the coeff of kinetic friction is 0.20. How long is the friction force exerted on the bock.

A block weighing 20 N rests on a horizontal surface. The coeff of static friction b/w the block and surface is 0.04 and the coeff of kinetic friction is 0.20. How long is the friction force exerted on the bock.

Grade:12

1 Answers

askIITianexpert IITDelhi
8 Points
11 years ago

As the Q say ,block is resting on the horizontal suraface.So unless some external horizontal force is applied on it there is no

frictional force on it.To move the block min. hor. force required would be equal to maximum static frictional force=(coeff. of st. fr.)*(normal reaction N exerted by surface on the block)

where N=weight of the block=mg

Until applied force is greater tha above block will not move.At any time frictional force on the block = force applied(upto maximum static frictional force ).

After that as block gets in motion kinetic frictional force will oppose the motion.Here again max. kin. fr. force=(coeff. of kin. fr.)*N

As applied force is already greater than this, the max. kin. fr. force is the only opposing force to the same & it will be so as long as applied force>max. kin. fr.force.Whenever it became equal or less block will stop & to get it in motion again you have to apply force>max. st. fr. force as stated above.

 

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