Verified Answer :

MathPhys

Answer:

0.83 J

Explanation:

According to the work-energy theorem, the work (W) done to stretch the spring is equal to the change in its elastic potential energy (ΔEE). This elastic energy (EE) is equal to half the spring constant (k) times the square of the displacement (x) from its equilibrium position.

W = ΔEE

W = ½ kx₂² − ½ kx₁²

W = ½ k (x₂² − x₁²)

Given that k = 23 N/m, x₁ = 0.24 m, and x₂ = 0.24 m + 0.12 m = 0.36 m, the work done to stretch the spring is:

W = ½ (23 N/m) ((0.36 m)² − (0.24 m)²)

W = 0.828 J

Rounded to two significant figures, the work done to stretch the spring is 0.83 J.

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