A block of mass 1 KG lies on a horizontal surface in a truck.The coefficient of a static friction between the block and the surface is 0.6.If the acceleration of the truck is 5 meter per second square,the frictional force acting on the block is
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377
mass of the block = 1kg
acceleration of the truck = 5 m/s²
pseudo force acting on the block = ma = 1×5 = 5N
coefficient of static friction = 0.6
maximum value of static friction = μmg = 0.6×1×10 = 6N
Friction is a self adjusting force. It can vary from 0 to its maximum value when a force acts on it. So when a force acts on a body, friction starts acting in the opposite direction. And movement of the body will not be there untill the force acting on it exceeds the maximum friction.
Here the force acting on it is 5N which is less than maximum friction(6N). So the friction will be 5N.
Note: maximum friction is 6N. friction is 5N.
acceleration of the truck = 5 m/s²
pseudo force acting on the block = ma = 1×5 = 5N
coefficient of static friction = 0.6
maximum value of static friction = μmg = 0.6×1×10 = 6N
Friction is a self adjusting force. It can vary from 0 to its maximum value when a force acts on it. So when a force acts on a body, friction starts acting in the opposite direction. And movement of the body will not be there untill the force acting on it exceeds the maximum friction.
Here the force acting on it is 5N which is less than maximum friction(6N). So the friction will be 5N.
Note: maximum friction is 6N. friction is 5N.
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43
F=8(0.42)×9.8(úmg)
=32.928
F=2(0.42)×9.8
=8.232
F=6(1.50)=9
F-41.23=9
F=50.23(51)
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