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Answer:
For an individual ( not a repair station) to conduct a complete 100 hour inspection on an aircraft and approve it for return to service requires a mechanic certificate with and powerplant and airframe ratings
Explanation:
Because these are important certificates issued by the FAA giving the mechanic authority to inspect to inspect an aircraft and approve its return to services
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Answer:
a) K_e = 0.1225 J, b) U = 1.96 J, c) v = 0.99 m / s
Explanation:
Let's use the simple harmonium movement expression
y = A cos (wt + Ф)
indicate that the amplitude is
A = 0.05 m
as the system is released, the velocity at the initial point is zero
v = dy / dt
v = - A w sin (wt + Ф)
for t = 0 s and v = 0 m/s
0 = - A w sin Ф
so Ф = 0
the expression of the movement is
y = 0.05 cos wt
The total energy of the system is
Em = ½ k A²
let's use conservation of energy
starting point. Spring if we stretch and we set the zero of our system at this point
Em₀ = K_e + U
Em₀ = 0
final point. When weight and elastic force are in balance
Em_f = K_e + U
Em_f = ½ k y² + m g (-y)
energy is conserved
Em₀ = Em_f
0 = ½ k y² + m g (-y)
k = 2mg / y
k = 2 4.00 9.8 / 0.050
k = 98 N / m
a) maximum elastic energy
K_e = ½ k A²
K_e = ½ 98 0.05²
K_e = 0.1225 J
b) the maximum gravitational energy
U = m g y
U = 4.00 9.8 0.05
U = 1.96 J
c) The maximum kinetic energy occurs when the spring is not stretched
U = K
mg h = ½ m v²
v = √2gh
v = √( 2 9.8 0.05)
v = 0.99 m / s
d) energy at any point
Em = K + U
