Respuesta :
Answer:
Therefore maximum stretch is y2 = 32.36 m
Explanation:
In this problem let's use the initial data to find the string constant, let's apply Newton's second law when in equilibrium
[tex]F_{e}[/tex] - W = 0
k Δx = mg
k = mg / Δx
k = 80 9.8 / (30-20)
k = 78.4 N / m
now let's use conservation of energy to find the velocity of the body just as the string starts to stretch y = 20 m
starting point. When will you jump
Em₀ = U = mg y
final point. Just when the rope starts to stretch
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
mg y = ½ m v²
v = √ 2g y
v = √ (2 9.8 20)
v = 19.8 m / s
now all kinetic energy is transformed into elastic energy
starting point
Em₀ = K = ½ m v²
final point
Em_{f} = [tex]K_{e}[/tex] + U = ½ k y² + m g y
Emo = Em_{f}
½ m v² = ½ k y² + mgy
k y² + 2 m g y - m v² = 0
we substitute the values and solve the quadratic equation
78.4 y² + 2 80 9.8 y - 80 19.8² = 0
78.4 y² + 1568 y - 31363.2 = 0
y² + 20 y - 400 = 0
y = [- 20 ±√ (20 2 +4 400)] / 2
y = [-20 ± 44.72] / 2
the solutions are
y₁ = 12.36 m
y₂ = 32.36 m
These solutions correspond to the maximum stretch and its rebound.
Therefore maximum stretch is y2 = 32.36 m
