a block of mass 10 kg is placed on a rough horizontal surface having coefficient of friction of 0.5 the minimum force required to move the block is
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Answered by
2
Answer:
F > 49 N
Step-by-step explanation:
Friction acting on it will be = uN
u=Coefficient of friction
N= Normal
Therefore Friction= umg
=0.5×10×9.8 =49 N
Therefore Force, let it be 'F' acting on it must be greater than 49 N.
Answered by
0
Answer:
answer is :20√5 because Fmin. is asked
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