A 20-Kg child is on a swing attached to 3.0 m-long chains. The child swings back and forth, swinging out to a 60-degree angle. (This is the angle that the chains make with the vertical.) What is the childs maximum speed on the swing

 v = 29.4 m / s

Explanation:

For this exercise we can use the conservation of mechanical energy

Lowest starting point.

          Em₀ = K = ½ m v²

final point. Higher

          = U = m g h

Let's use trigonometry to lock her up

          cos 60 = y / L

          y = L cos 60

Height is the initial length minus the length at the maximum angle

           h = L - L cos 60

           h = L (1- cos 60)

energy is conserved

         Em₀ = Em_{f}

          ½ m v² = mgL (1 - cos 60)

         v = 2g L (1- cos 60)

let's calculate

          v² = 2 9.8 3.0 (1- cos 60)

          v = 29.4 m / s

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.