A 19.0 kg child on a 1.00 m long swing is released from rest when the ropes of the the swing make an angle of 32.0° with the vertical. (a) Neglecting friction, find the child's speed at the lowest position. L - Lcostheta = h 1 -1cos32 = .15 mgh = 1/2mv^2 (19)(9.8)(.15) = 1/2(19)v^2 27.93 = 9.5v^2 v = 1.7 m/s (b) If the speed of the child at the lowest position is 1.50 m/s, what is the mechanical energy lost due to friction? J I do not understand this question at all but i think its like this: KE = 1/2(19)(1.5)^2 then i would get mgh and subtract that by KE? KE = 21.38 27.93 - 21.38 = 6.55 J? please help me out i seriously dont understand this question.