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A block of mass 0.1kg is held against a wall by applying horizontal force of 5N on the block.if the coefficient of friction of the block and wall is 0.5 then the magnitude of the frictional force acting on the block is?

Aryaman , 7 Years ago
Grade 12
anser 3 Answers
Arun

Last Activity: 7 Years ago

Dear student
 
Since the block is held against a wall, the coefficient of friction will be equal to the weight of the block.
 
Hence
mu = m*g = 0.1 *9.8 = 0.98 N

Disha Sharma

Last Activity: 5 Years ago

The frictional force will be directed upwards as shown as gravity will be assisting its motion and the frictional force will be due to 5 N force normal to the block towards the wall.
Hence max. static frictional force: F = 0.5 x 5 = 2.5 N
FOrce due to mass of block= mg = 0.1 x 10 = 1 N
Since it is less than max static frictional force hence frictional force on the block will be 1 N

ankit singh

Last Activity: 4 Years ago

Limiting frictional force, fl=μsN=0.5×5=2.5N. tending to produce relative motion is the weight (W) of the block which is less than fl. Therefore, the frictional force is equal to the weight, the magnitude of the frictional force f has to balance the weight 0.98 N acting downwards. Therefore the frictional force =0.98N

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