A 31.0-kg child on a 3.00-m-long swing is released from rest when the ropes of the swing make an angle of 27.0° with the vertical. (a) Neglecting friction, find the child's speed at the lowest position.
(b) If the actual speed of the child at the lowest position is 2.20 m/s, what is the mechanical energy lost due to friction?

a)  v = 2.53 m / s, b)  ΔEm = -24.32 J

Explanation:

a) For this exercise we can use the conservation of energy,

starting point. Highest point

           Em₀ = U = m g h

final point. Lower of the trajectory

           Em_f = K = ½ m v²

as there is no friction energy is conserved

           Em₀ = Em_f

           mgh = ½ m v²

           v² = 2gh

we are used trigonometry to find the height

           h = L - L cos tea

we substitute

           v =

let's calculate

           v =

           v = 2.53 m / s

b) The loss of mechanical energy due to friction

          ΔEm = Em_f -Em₀

          ΔEm = ½ m v² - m g h

          ΔEem = ½ m v² - m g L (1-cos θ)

let's calculate

          ΔEm = ½ 31.0  2.20² - 31  9.8  3.00 (1-cos 27)

           ΔEm = 75.02 - 99.34

           ΔEm = -24.32 J

the negative sign indicates that power has been requested

the correct choice is a .

do not post any more test or exam questions on

ice to water 1 water to vapour 2 latent heat