Respuesta :
Answer:
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 = [tex]\sqrt{2gL (1- cos \theta)}[/tex]
let's calculate
v = [tex]\sqrt { 2 \ 9.8 \ 3.00 \ ( 1- cos \ 27) }[/tex]
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
