Questions: After the block is released, in which of the following sequences of positions is the speed of the block ordered from fastest to slowest? A block released from rest at position A slides with negligible friction down an inclined track, around a vertical loop, and then along a horizontal portion of the track, as shown above. The block never leaves the track. (A) B C D E (B) B E C D (C) D C E B (D) E a C D (E) E D C B

Solution Steps

The problem involves a block sliding down an inclined track, around a vertical loop, and then along a horizontal portion. We need to determine the sequence of positions where the block's speed is ordered from fastest to slowest.

Since the block is released from rest and friction is negligible, we can use the conservation of mechanical energy. The total mechanical energy (potential energy + kinetic energy) remains constant throughout the motion.

• At the highest point (position A), the block has maximum potential energy and zero kinetic energy.
• As the block descends, potential energy is converted into kinetic energy, increasing its speed.
• The speed is highest at the lowest point of the track, where potential energy is minimal.

• Position B: The block is moving down the incline, gaining speed.
• Position C: The block is at the bottom of the loop, where it has the maximum speed due to the lowest potential energy.
• Position D: The block is moving up the loop, losing speed as potential energy increases.
• Position E: The block is on the horizontal portion, maintaining speed but not as fast as at the bottom of the loop.

Based on the analysis:

• The fastest speed is at position C (bottom of the loop).
• The next fastest is at position B (descending the incline).
• Then position E (horizontal portion).
• The slowest is at position D (ascending the loop).

Final Answer