A small block of mass \(m \) slides along the frictionless loop-the-loop track . If it starts at \(A\) at height \(h=5R\) from the bottom of the track then what will be the resulting force acting on the track at \(B\)?


Let the velocity at \(B\) be \(v\).
kinetic energy \(\frac{1}{2}mv^2\)=potential energy lost \(mg(5R-R)\)
$$ \frac{mv^2}{R}=8mg$$
This is the centrigular force acting horizontally,
The weight acts vertically down.
Hence, the resultant force
$$ F=\sqrt{(8mg)^2+(mg)^2}=\sqrt{65}mg$$