Respuesta :
assuming the reference line to measure the height for gravitational potential energy lying at the equilibrium position
m = mass attached to the spring = 10.00 kg
k = spring constant of the spring = 250 N/m
h = height of the mass above the reference line or equilibrium position = 0.50 m
x = compression of the spring = 0.50 m
v = speed of mass = 2.4 m/s
A = maximum amplitude of the oscillation
v' = speed of mass at the maximum amplitude location = 0 m/s
using conservation of energy between the point where the speed is 2.4 m/s and the highest point at which displacement is maximum from equilibrium
kinetic energy + spring potential energy + gravitational potential energy = kinetic energy at maximum amplitude + spring potential energy at maximum amplitude + gravitational potential energy at maximum amplitude
(0.5) m v² + m g h + (0.5) k x² = (0.5) m v'² + m g A + (0.5) k A²
inserting the values
(0.5) (10) (2.4)² + (10) (9.8) (0.50) + (0.5) (250) (0.50)² = (0.5) (10) (0)² + (10) (9.8) A + (0.5) (250) A²
109.05 = (98) A + (125) A²
A = 0.62 m
