block of mass m slides down a frictionless incline as
The block is released from height h above the bottom of the circular loop. (a) Find the force exerted on the block by the inclined track at point A. (Hint: Consider the Newton’s 2nd law in the radial direction and you may use polar coordinates for this circular part of the path to relate the acceleration and velocity components.) (b)Find the force exerted on the block by the inclined track at point B. (c)Find the speed of the block at point B. (d)What is the maximum height that the block can reach after leaving the track? (e)Find the distance between point A and the point that the block land on ground level?
To find the force exerted on the block by the inclined track at point A, we will use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.
(a) First, let's analyze the forces acting on the block at point A:
1. Weight (mg): This force acts vertically downward and is given by m multiplied by the acceleration due to gravity (g).
Since the block is sliding down a frictionless incline, there is no friction force present. Therefore, the only force acting on the block is its weight.
The force exerted by the inclined track at point A can be found using polar coordinates. In polar coordinates, we can decompose the weight force into two components: one along the radial direction (Fn) and one along the tangential direction (Ft). The radial component (Fn) provides the force needed to keep the block moving in a circle.
At point A, the radial component (Fn) is equal to the weight force (mg) because all of the weight force is directed towards the center of the circular loop.
Therefore, the force exerted on the block by the inclined track at point A is Fn = mg.
(b) Moving on to point B, the force exerted on the block by the inclined track changes due to the change in direction. At point B, the radial component (Fn) is no longer equal to the weight force (mg).
To find the force at point B, we need to consider the radial and tangential components separately:
1. The radial component (Fn) provides the force needed to keep the block moving in a circle and is perpendicular to the incline.
2. The tangential component (Ft) provides the force that accelerates the block along the incline.
At point B, the radial component (Fn) is zero because it is directed orthogonal to the incline. Therefore, the force exerted on the block by the inclined track at point B is solely the tangential component (Ft).
(c) To find the speed of the block at point B, we can use the conservation of energy. The total mechanical energy at point B is equal to the sum of the potential energy at point A and the kinetic energy at point B:
Potential energy at point A: mgh
Kinetic energy at point B: (1/2)mv^2
Given that there is no dissipative force (e.g., friction), the total mechanical energy remains constant. Therefore, we can equate the potential energy and kinetic energy expressions and solve for the speed (v) at point B.
mgh = (1/2)mv^2
Solving for v, we find v = √(2gh).
(d) To determine the maximum height that the block can reach after leaving the track, we need to consider the conservation of energy again. The block reaches its maximum height when its kinetic energy is zero. So, we equate the initial gravitational potential energy (mgh) to the final gravitational potential energy (mgΔh):
mgh = mgΔh
Simplifying, we find Δh = h/2. Therefore, the maximum height that the block can reach after leaving the track is h/2.
(e) Finding the distance between point A and the point where the block lands on the ground level requires the information regarding the radius of the circular loop or any other relevant data about the circular part of the path. Unfortunately, that information is not provided in the question, making it impossible to calculate the distance between point A and the landing point on the ground level without additional information.