A block of mass 4 Kg is placed in contact with the front vertically surface of the Lorry. the coefficient of friction between the vertically up surface and block is 0.8. the lorry is moving with an acceleration of 15 metres per second. the force of friction between the Lorry and the block is

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Solution

Mass of the block = 4kg
acceleration of the truck = 15 m/s²
pseudo force acting on the block = ma = 4×15 = 60N
coefficient of static friction = 0.8
maximum value of static friction = μmg = 0.8×4×10 = 32N

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 60N which is greater than maximum friction(32N). So the friction will be 32N.
Note: maximum friction is 32N. friction is 32N.

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