In a children park an inclined plane is constructed with an angle of incline `45^0` in the middle part figure. Find the acceleration of a boy sliding on it if the friction coefficient between the cloth of the boy and the incline is 0.6 and g=10 m/`s^2`.

Answer

`2.828m//s^(2)`

Explanation

Here, `theta = 45^(@), a = ? mu = 0.6, g = 10 ms^(-2)`
`a = g (sin theta - mu cos theta)`
`=10 (sin 45^(@) - 0.6 cos 45^(@))`
`= 10 ((1)/(sqrt2) - (0.6)/sqrt2) = 10xx(0.4)/(sqrt 2) = (4)/(sqrt(2))`
`a = (4sqrt2)/(2) = 2 xx 1.414 = 2.828ms^(-2)`.