A body of mass m is placed on a platform of mass M ( `m lt M`), moving with a velocity v. If the coefficient of friction between the platform and the mass is `mu` then for how long will the body continue to slide on the platform and what distance will it cover during that time?

Explanation

When the platform moves the body on it tends to slide backwards. So the frictional force on the body acts in the direction of motion of the platform. It is given by , f = `mu`mg.
`:.` Its acceleration a = `(f)/(m) = mu g`
When the velocity of the body becomes v it no longer slides over the platform.
We know v = u+at `" ""or", t = (v+u)/(a)`
Here u=0, a = `mu g :. t = (v)/(mu g)`
Again `v^(2) = u^(2) +2 as " ""or", s = (v^(2)-u^(2))/(2a) = (v^(2))/(2mu g)`
`:.` The body will slide for a time of `(v)/ (mu g)`, and in this interval of time it will cover a distance of `(v^(2))/(2mug).`