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

A small block of mass m slides along a smooth track as shown in the fi

1.	A small block of mass m slides along a smooth track as shown in the fig.(i) If it starts from rest at P, what is the resultant force acting on it at Q ?(ii) At what height above the bottom of the loop should the block be released so that the force it exerts against the track at the top of the loop equals its weight ?
Answer» Solution :By conservation of mechanical energy between points P and Q. `mg(5R)=mgR + (1)/(2) mv^(2)` i.e., `v= SQRT(8gR)` Now in case of circular motion. N (or T) `= (mv^(2))/(R )+mg cos theta` And as at Q, `theta = 90^(@)` `N= (mv^(2))/(R )=(m(8gR))/(R )=8mg` So resultant force on m at Q. `F = sqrt((8mg)^(2)+(mg)^(2))=(sqrt(65))mg` (ii) At HIGHEST point `N=(mv^(2))/(R )-mg"" (as theta = 180^(@))` But according to given problem N = mg so `(mv^(2))/(R )=mg + mg`, i.e., `v=sqrt(2gR)` If for achieving it h. the height, by conservation of mechanical energy again `MGH. =(1)/(2)mv^(2)+mg(2R)` or `h.=2R+(v^(2))/(2g)=2R+(2gR)/(2g)=3R`