Respuesta :
Answer:
k = 930 N / m
Explanation:
For this problem we will solve it in parts, let's start with the conservation of mechanical energy
Starting point. Lower
Em₀ = [tex]K_{e}[/tex] + U₁
Final point. Higher
[tex]Em_{f}[/tex]= U₂
as there is no friction, energy is conserved
Em₀ = Em_{f}
½ m v² + mg y₁ = m g y₂
where y₁ is the initial height of y1 = -0.5 m and y² the final height y² = 15 m
Let's find speed when getting out of the sling
v = √ (2g (y₂-y₁))
let's calculate
v = √[2 9.8 (15 - (-0.5))]
v = 17.43 m / s
Now we can use Newton's second law.
The force applied by the sling is in the direction of movement (inclined) and the weight is in the vertical direction.
X axis
Fₓ = m aₓ
in the problem they indicate that the direction of the velocity at the end of the sling is the same direction of the force,
F_{e} cos θ = m a cos θ
let's replace the elastic force
k Δx = m a
Y axis
F_{y} - W = m a_{y}
k Δx sin θ - m g = m a sin θ
let's write
k Δx = m a (1)
k Δx sin θ - m g = m a sin θ
Now let's use kinematics to find the acceleration in the sling, the direction of these accelerations ta in the direction of elongation
v² = v₀² +2 a Δx
as the system starts from rest v₀ = 0
a = v² / 2Δx
a = 17.43² / (2 1.4)
a = 108.5 m / s²
we substitute in equation 1
k = m a / Δx
k = 12 108.5 / 1.4
k = 930 N / m
