Question

at what position does the rock have the greatest kinetic energy? a d e b

Explanation:

Step1: Recall Kinetic Energy (KE) Concept

Kinetic energy depends on speed ($KE = \frac{1}{2}mv^2$). A rock falling (or moving down) gains speed due to gravity, losing potential energy (PE) and gaining KE. At the lowest point (just before hitting the ground or the lowest position in the path), speed is maximum.

Step2: Analyze Positions

• Position A: Maybe the start (low, but if the rock is at rest here, KE is low).
• Position B: Climbing up, speed low (maybe moving up, KE not max).
• Position C: At the top, just released? Speed 0 initially, KE 0.
• Position E (assuming E is the lowest point, like near the bottom of the hill or just before landing): As the rock falls, it accelerates due to gravity. At the lowest point (E), it has the maximum speed (before hitting the ground, where friction/impact would reduce it). So at E, KE is greatest (since $KE \propto v^2$, and v is maximum here). Wait, looking at the diagram, E is the lowest position (the bottom of the falling path). So when the rock reaches E, it's moving fastest (after accelerating down from C/D). So the position with greatest KE is E.

Answer:

E (the lowest position, e.g., the bottom of the falling path shown in the diagram)