Question

A small block slides without friction along a track toward a circular loop. The block has more than enough speed to remain firmly in contact with the track as it goes around the loop. The magnitude of the blocks acceleration at the top of the loop is

Answer 1

The magnitude of the block's acceleration at the top of a frictionless circular loop consists of the centripetal acceleration required to maintain its circular path and the acceleration due to gravity. At the top of the loop, the total acceleration is thus equal to the acceleration due to gravity, approximately 9.8 m/s² downwards.

Step-by-step explanation:

You are asking about the magnitude of the block's acceleration at the top of a frictionless circular loop. In physics, acceleration of an object moving in a circle can be described by the centripetal acceleration formula a = v^2/r, where v is the velocity of the object and r is the radius of the circle. At the top of the loop, besides the centripetal acceleration, the block is also experiencing acceleration due to gravity, acting downwards. So, the total acceleration is a combination of the centripetal acceleration needed to stay in a circular path, and the acceleration due to gravity.

To understand this concept, we can use Newton's second law in the context of circular motion. When the small block is at the top of the loop, the only force acting towards the center of the circle, which provides the centripetal force, is the weight of the block (mg, where 'm' is the mass of the block and 'g' is the acceleration due to gravity). Therefore, at the top of the loop, the centripetal force equals the gravitational force (Fc = mg), and thus the centripetal acceleration is equal to the acceleration due to gravity (ac = g). This results in the block's total acceleration at the top of the loop being g, or approximately 9.8 m/s² downwards.