Answer: 0.6u 2sin 2 (theta)/(2g).
- A 0.6u<sup>2sin</sup><sup>2</sup>(theta)/(2g)
- B 0.6u<sup>2</sup>/(2g)
- C 0.6u<sup>2sin</sup><sup>2</sup>(theta)/g
- D 0.3u<sup>2sin</sup><sup>2</sup>(theta)/g
Correct answer: A. 0.6u<sup>2sin</sup><sup>2</sup>(theta)/(2g)
Explanation: Total E = (1/2)mu<sup>2.</sup> PE = mgh = 60% of E = 0.3mu<sup>2.</sup> h = 0.3u<sup>2</sup>/g. But vertical KE component: PE = 0.6 x (1/2)m(u sin theta)<sup>2</sup> = mgh. h = 0.6u<sup>2sin</sup><sup>2</sup>(theta)/(2g) = 0.3u<sup>2sin</sup><sup>2</sup>(theta)/g.
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.