A block of mass 2 kg is placed on the fl...

A block of mass 2 kg is placed on the floor . The coefficient of static friction is `0.4` . If a force of `2.8` N is applied on the block parallel to floor , the force of friction between the block and floor is : (Taking g = `10 m//s^(2)`)

2.8 N

8 N

2 N

zero

Text Solution

Verified by Experts

The correct Answer is:

Force of friction , f = `mu "mg" = 0.4 xx 2 xx 10 = 8 N`
As F `lt` f , the body will be in static equilibrium .
`therefore " " f = 2.8 N`

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