A block of mass 10kg is placed on a rounh horizontal surface of coefficient of friction 0.5 . The minimum force required to move the block is(g=10m/s)
'Answer is not 50N'
Please solve.
Answers
Answered by
0
Answer:
Explanation:
Mass of block = 10 kg
Force acting on block = mg
( Which is equal to the normal force )
Therefore maximum frictional force =mu x mg
mu - coefficient of friction = 0.5
Therefore minimum friction required to start moving the block is mu x mg
= 0.5 x 10 x g
= 5g
= 50 ( g = 10 )
Attachments:
Similar questions