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a man of mass 60kg standing on a block of mass 40kg kept on ground.the co efficient of friction between the feet of man and the block is 0.3 and that between block and ground is 0.2.if the man pulls the one end string with 125N force and another end of string is connected to block through a pulley fixed to the wall in front,thena)block will slide on ground b)man and block move with accEleration 0.5c)friction acting between man and block is 40Nd)the friction acting between man and block is 180..multi answer type question ............please with explain free body diagrams

akash kashyp , 14 Years ago
Grade Upto college level
anser 2 Answers
Badiuddin askIITians.ismu Expert

Last Activity: 14 Years ago

Dear akash

for slipping take place between man and block so T must be greater than  0.3*N =0.3*60 *g=180

but T is less than 180N so slipping not take place

 

6369-84_7662_untitled.JPG

let the whole accembly will move eith acceleration a

so

2T-(60+40)*.2*g =(60+40)a

250-200=100a

a=.5

let the friction between man and block is f

T+f=60*.5

  f=30-125=-95 N(opposite direction as assumed)

so aption a and b are correct

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Snehankur Mondal

Last Activity: 6 Years ago

For the entire system to have an acceleration,the frictional force must be less than the tensional force. But ut according to the calculations it is not the case here.
We find the frictional force to be 180N whereas the tensional force is 125N.
Hence there will be no acceleration in the entire system.
Therefore we can conclude that a =0 .
So if there is no acceleration  the entire system will be an equilibrium.
Therefore tensional force(T) + frictional force(f)=net force acting on the body.
But we have found that
a=0
therefore f=T
i.e. frictional force acting between he foot of the man and the block is equal to tensional force which is equal to 125 N
Similarly in case of the block the frictional force acting between the block and the ground is 2T equal to 250 N.

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