A block of mass `10 kg` is placed on a horizontal surface as shown in the figure and a force `F = 100 N` is applied on the block as shown, in the figure. The block is at rest with respect to ground. If the contact force between block and ground is `25n` (in Newton) then value of n is : `(Take g = 10 m//s^(2))`

Answer

4

Explanation

A block of ……………
`F sin 30^(@) = 50N`
So normal is `50 N`
and `f_(max) = mu N = 100 N lt F cos 30^(@)`
So block is in equilibrium

`vecR+vecF+vec(Mg) = 0`
`vecR= -(vecF+vec(Mg))`
`R = 100 N`