. A particle is projected up along a rough inclined plane of inclination `45^(@)` with the horizontal. If the coefficient of friction is 0.5, the acceleration is

Options

A. `(g)/(2)`
B. `(g)/(2sqrt2)`
C. `(3g)/(2sqrt2)`
D. `(g)/(sqrt2)`

Answer

(c)

Explanation

`theta =45^(@)`
`mu=0.5`
Acceleration, `a=g(sintheta+mucostheta)`
`=g(sin45^(@)+0.5cos45^(@))`
`=g((1)/(sqrt2)+(0.5)/(sqrt2))`
`=(3g)/(2sqrt2)`