A vertical spring with a force constant of 26 N/m has a relaxed length of 35 cm. A 200 g mass is then attached to the end of the spring and is allowed to come to rest. (a) Find the length of the stretched spring.
(b) How much elastic potential energy is stored in the spring?

I just need the work please the answers is
(a) 45.2 J
(b) 0.0739

(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 = L + x

∴ L ≈ 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

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