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)/(g(sin theta-mucos theta))) rArr (T_(1))/(T_(2))=sqrt((sin theta-mu cos theta)/(sin theta))`
`T_(2)=T_(1)sqrt((sin theta)/(sin theta-mu cos theta))`
`=2sqrt((sin45^(@))/(sin45^(@)-(0.5)cos45^(@)))`
`=2sqrt(((1//sqrt2))/((1//sqrt2)-(0.5)(1//sqrt2)))=2xxsqrt2=2.828s`