# Mechanics prob: energy conservation with a swing pls help... ?

My answer for the lowest point works.

But I'm having trouble doing the highest point one..?

Explanation with answer pls :/

### 2 Answers

- husoskiLv 73 weeks agoBest answer
Is this a trick question? The ball *starts* at the highest point in its swing, and the speed is zero there!

The potential energy at a distance h below the rest position, relative to that starting position, is -mgh, right? So conservation of energy and the work-energy theorem say that all of that P.E. lost goes to kinetic energy:

mgh = (1/2)mv²

v = √(2gh)

At the bottom of both circles, h = L = 1.080 m, so the speed is:

v = √(2 * 9.81 * 1.080) ~~ 4.60 m/s

At the top of the small circle, h = d - r = d - (L - d) = 2d - L = (2*0.756 + 1.080) m

h = 0.432 m

At that position, the speed is:

v = √(2 * 9.81 * 0.432) ~~ 2.91 m/s

- Born YesterdayLv 73 weeks ago
108 - 75.6 = 32.4 = r (from image)

108 - 2×32.4 = 43.2 cm (distance below

original starting point)

From energy conservation:

m/2×v^2 = mgh

v = (2gh)^.5 = (19.6×.432)^.5 = 2.91m/s

Check: (19.6×1.08)^.5 = 4.6 m/s

Update if you need more, and please choose

a best answer from among those you get.

Ohh I see, but then how is it that at the top of of the small circle we're equaling Mechanical Energy = PE? Aren't we supposed to equal Mechanical Energy = PE + KE since there's kinetic energy at the top as well?