A block is moving on an inclined plane making an angle `45^(@)` with the horizontal and the coefficient of friction is `mu` . The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down . If we define `N = 10 mu` , then N is :

Answer

5

Explanation

`F_(1) = 3F_(2)`
mg (sin `45^(@) + mu` cos `45^(@)`) = 3 mg (sin `45^(@) - mu "cos" 45^(@))`
`(1)/(sqrt2) ( 1+ mu) = (3)/(sqrt2) (1 - mu)`
`mu = 0.5`
`therefore " " N = 10 mu`
= `10 xx 0.5`
N = 5