Respuesta :
Answer:
(a) The length of the stretched string is approximately 42.53846 cm
(b) The elastic potential energy stored in the string is approximately 0.073877 Joules
Explanation:
(a) The given parameter are;
The force constant of the vertical spring, k = 26 N/m
The relaxed length of the spring, L = 35 cm = 0.35 m
The mass attached to the end of the spring, m = 200 g = 0.2 kg
By Hooke's law, we have;
F = k·x
Where;
F = The applied force on the sting = The weight of the attached mass, W
∴ F = W
The weight of the attached mass, W = m × g
Where;
g = The acceleration due to gravity ≈ 9.8 m/s²
Therefore, F = W = 0.2 kg × 9.8 m/s² = 1.96 N
From Hooke's law, we have;
x = F/k = 1.96 N/(26 N/m) ≈ 0.0753846 m
The extension of the stretched spring, x ≈ 0.0753846 m
The length of the stretched string, L[tex]_{stretched}[/tex] = L + x
∴ L[tex]_{stretched}[/tex] ≈ 0.35 m + 0.0753846 m ≈ 0.4253846 m
0.4253846 m = 42.53846 cm
The length of the stretched string ≈ 42.53846 cm
(b) The elastic potential energy stored in the string, U = 1/2·k·x²
By substituting the known values, we get;
U = 1/2 × 26 N/m × (0.0753849 m)² ≈ 0.073877 Joules
