Which option gives an
1 answer:
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Answer:
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
Answer:
x = 76.5 m
Explanation:
Let's use Newton's second law at the point of contact between the wheel and the floor.
fr = m a
fr = miy N
N-W = 0
N = W
μ mg = m a
a = miu g
a = 0.600 9.8
a = 5.88 m / s²
Having the acceleration we can use the kinematic relationships to find the distance
² = v₀² + 2 a x
= 0
x = -v₀² / 2 a
Acceleration opposes the movement by which negative
x = - 30²/2 (-5.88)
x = 76.5 m