A block of mass 8 kg is sliding on a surface inclined at an angle of `45^(@)` with the horizontal . Calculate the acceleration of the block . The coefficient of friction between the block and surface is `0*6` ( Take g = `10 m//s^(2)`)

Explanation

Here ` m = 8 kg , theta = 45^(@) , a =? , mu = 0* 6 `
Here , `tan theta 45^(@) = 1 gt mu (=0*6)`

So, body starts moving and acceleration of the body .
` a = g(sin theta - mu cos theta ) `
`:. a = 10 (sin 45^(@) - 0*6 cos 45^(@)) `
` = 10 [ 0*707 - 0*6 xx 0*707 ]`
` = 10 xx 0* 707 ( 1- 0*6)`
` 10 xx 0* 707 xx 0*4 `
` 2* 828 m//s^(2)`