A small block of mass `m` slides along a smooth frictionless track as shown.
(a) If it starts from rest at `P`, what is the resultant force acting on it at `Q`?
(b) At what height above the bottom should the block be released so that the force it exerts against the track at the top of the loop equals its weight?

Explanation

(a) when the block reaches at `Q`, its speed is `v`.
`v^(2)=u^(2)+2gh=0+2g(5R-R)=8gR`
`N=(mv^(2))/(R )=8mg`
weight of the block `W=mg`
Resultant force on the block when it is at `Q` is
`sqrt(N^(2)+W^(2))=sqrt((8mg)^(2)+(mg)^(2))`
`=sqrt(65)mg`
(b) when the block is at `H,N'+mg=(mv_(0)^(2))/(R )`,
Given `N'=mg`
`2mg=(mv^_(0)^(2))/(R )impliesv_(0)^(2)=2gR`
`v_(0)^(2)u^(2)+2g(h-2R)`
`2gR=0+2g(h-2R)`
`h=3R`