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A block of mass 1kg lies on a horizontal surface in a truck the co efficient of a static friction between the surfaces 0.6 if the acceleration of the trunk

is 5m\s the frictional force acting on the block is



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Mass of block = 1 kg
acceleration of truck = 5 m/s²
pseudo force acting on block = 1×5 = 5N
maximum friction force that can act on block = μmg = 0.6×1×9.8 = 5.88 N

Since force acting on block is less than the maximum friction force, the friction force acting on it is 5N.
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