Question

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.

Answer 1

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 )