A body moving on the ground with a velocity of `15m//s` comes to rest after covering a distance of `25m` If acceleration due to gravity is `10m//s^(2)` Calculate the coefficient of friction between the ground and the body .

Answer

`0.45`

Explanation

Here, `u = 15m//s, upsilon = 0 s = 25 m`
`g = 10m//s^(2) mu =` ?
From `upsilon^(2) - u^(2) = 2 as`
` a = (upsilon^(2) -u^(2))/(2 s) =(0^(2) - 15^(2))/(2 xx 25) = - 4.5m//s^(2)`
`mu = (a)/(g) = (4.5)/(10) = 0.45` .