Answer: sqrt(2g h(1 - mu_k cot(theta))).
- A sqrt(2gh)
- B sqrt(2g(h - mu_k x h/tan(theta)))
- C sqrt(2g(h - mu_k h cos<sup>2</sup>(theta)/sin(theta)))
- D sqrt(2g h(1 - mu_k cot(theta)))
Correct answer: D. sqrt(2g h(1 - mu_k cot(theta)))
Explanation: Energy at bottom = mgh - friction work. Friction work = mu_k mg cos(theta) x (h/sin(theta)) = mu_k mgh/tan(theta). v = sqrt(2gh(1 - mu_k/tan(theta))).
As a pendulum swings, total mechanical energy (KE+PE) stays constant: at the highest points A and C, all energy is potential (zero speed); at the lowest point B, all of that energy has converted into kinetic energy (maximum speed).
Concept context
Work-energy theorem, potential energy, conservation of energy, collisions.