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) )= 2sec`
Time taken with friction is
`T_(2)=sqrt((2l)/(s(sin theta-mucostheta)))rArrT_(1)/T_(2)= sqrt((sin theta -mu cos theta)/(sin theta))`
`T_(2) T_(1)sqrt((sintheta)/(sin theta-mucostheta))`
`=2sqrt((sin45^(@))/(sin 45^(@)-(0.5)cos45^(@)))`
`=2sqrt(((1//sqrt2))/((1//sqrt2)-(0.5)(1//sqrt2))) = 2 xx sqrt2 = 2.828s` .