Respuesta :
Answer:
a) v = 18.86 m / s, b) h = 8.85 m
Explanation:
a) For this exercise we can use the conservation of energy relations.
Starting point. Like the compressed spring
Em₀ = K_e + U = ½ k x² + m g x
the zero of the datum is placed at the point of the uncompressed spring
Final point. With the spring if compress
Em_f = K = ½ m v²
how energy is conserved
Em₀ = Em_f
½ k x² + m g x = ½ m v²
v² = [tex]\frac{k}{m}[/tex] x² + 2gx
let's reduce the magnitudes to the SI system
m = 500 g = 0.500 kg
x = -45 cm = -0.45 m
the negative sign is because the distance in below zero of the reference frame
let's calculate
v² = [tex]\frac{900}{0.500}[/tex] 0.45² + 2 9.8 (- 0.45)
v = √355.68
v = 18.86 m / s
b) For this part we use the conservation of energy with the same initial point and as an end point at the point where the rock stops
Em_f = U = m g h
Em₀ = Em_f
½ k x²2 + m g x = m g h
h = ½ [tex]\frac{k}{g}[/tex] x² + x
let's calculate
h = [tex]\frac{1}{2} \ \frac{900}{9.8 } \ 0.45^2[/tex] - 0.45
h = 8.85 m
measured from the point where the spring is uncompressed
