A small block of mass m slides along a friction less loop. If it starts from rest at P, what is the resultant force acting on it at Q?

Since the loop is frictionless, theere are only two forces acting on the block: Gravity and normal force. Gravity pulls the block downward, and the surface of the track pushes out perpendicular to the track.The force of gravity on the block is fixed: mg. The direction of gravity is always downward.The normal force will vary, however, as is necessary to keep the block from falling through the track. When the track is...

Since the loop is frictionless, theere are only two forces acting on the block: Gravity and normal force. Gravity pulls the block downward, and the surface of the track pushes out perpendicular to the track.

The force of gravity on the block is fixed: mg. The direction of gravity is always downward.

The normal force will vary, however, as is necessary to keep the block from falling through the track. When the track is at an angle x with the ground, the force of gravity into the track will be mg cos x, and this will be exactly compensated by a normal force of mg cos x.

At the indicated point Q, the track appears to be vertical.
This means that x = pi/2, and thus cos x = 0; that is, the track does not push against the block at all, because the only force on the block is directly downward, parallel with the track.

Provided that the block has enough energy to reach that point at all (and from the diagram it looks as though it does), the only force acting on it at Q will be gravity, mg.