A small block of mass m slides along the frictionless loop track,if it starts from rest at point p,what us the resultant force acting on it at q?
angle of the slope
mg*sinangle is the F down or resultant force
To determine the resultant force acting on the block at point Q, we need to consider the forces acting on the block as it moves along the loop track. In this scenario, there are two main forces at play:
1. Gravitational Force (Weight):
The weight of the block, denoted as mg (where g is the acceleration due to gravity), is acting vertically downwards. At any point on the track, this force can be split into two components: the normal force (N) acting perpendicular to the track, and the force of gravity (mg) acting vertically downwards.
2. Centripetal Force:
As the block moves along the loop, it experiences a centripetal force that keeps it moving in a circular path. This force is directed towards the center of the loop and is responsible for the block's acceleration.
At point P, the block starts from rest, which means its initial velocity is zero. As it moves along the loop, the gravitational force (mg) provides the necessary centripetal force for the circular motion. Therefore, at point Q, the resultant force acting on the block is the only force present: the gravitational force, which acts radially towards the center of the loop.
To calculate this force precisely, you can use the following equation:
F_resultant = m * (v^2 / r)
Where:
- F_resultant is the resultant force acting on the block,
- m is the mass of the block,
- v is the velocity of the block,
- r is the radius of the loop at point Q.
Since the block starts from rest, its velocity at point Q will depend on the radius of the loop and the height difference between points P and Q. If you have these values, you can calculate the velocity and substitute it into the equation above to find the resultant force acting on the block at point Q.