A body of mass 'm' slides down a smooth inclined plane having an inclination of `45^(@)` with the horizontal . It takes 2 s to reach the bottom . If the body is placed on a similar plane having coefficient of friction `0.5` , then what is the time taken for it to reach the bottom ?

Explanation

Mass = m `theta = 45^(@) , mu= 0.5` Time taken by the body to reach the bottom without friction is
`T_(1) = sqrt((2l)/(g "sin" theta)) = 2 ` sec
Time taken with friction is
`T_(2) = sqrt((2l)/(g("sin" theta - mu "cos" theta)))`
`implies " " (T_(1))/(T_(2)) = T_(1) sqrt(("sin"theta - mu "cos" theta)/("sin" theta))`
`T_(2) = T_(1) sqrt(("sin" theta )/("sin" theta- mu "cos" theta))`
`=2 sqrt(("sin" 45^(@))/("sin" 45^(@) - (0.5)"cos" 45^(@)))`
`=2sqrt(((1//sqrt2))/((1//sqrt2) - (0.5)(1//sqrt2)))`
= `2 xx sqrt2 = 2.828` s