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How much does the elastic potential energy in the spring change?

User Brian Chen
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Final answer:

The elastic potential energy of a perfectly elastic spring with a spring constant of 4 N/cm increases by 0.54 J when its length increases from 23 cm to 26 cm.

Step-by-step explanation:

Change in Elastic Potential Energy

The change in elastic potential energy of a spring can be calculated using the formula for potential energy of a spring (PEs), which is PEs = 1/2 kx², where k is the spring constant and x is the displacement from the equilibrium position. For a perfectly elastic spring with a spring constant of 4 N/cm, and an unstretched length of 20 cm, when the length is increased from 23 cm to 26 cm, the displacement changes from 3 cm (0.03 m) to 6 cm (0.06 m).

Using the given spring constant and the displacement for each case, we calculate the initial potential energy when x = 3 cm as Uinitial = 1/2(4 N/cm)(3 cm)² = 0.18 J. Subsequently, the potential energy at a displacement of x = 6 cm is Ufinal = 1/2(4 N/cm)(6 cm)² = 0.72 J. The change in potential energy is then the difference, ΔPE = Ufinal - Uinitial = 0.72 J - 0.18 J = 0.54 J.

User Drk
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