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A spring with spring constant of 20 N/m is stretched 0.16 m from its equilibrium position. How much work must be done to stretch it an additional 0.063 m? Answer in units of J.

User Ohbrobig
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1 Answer

4 votes

Answer: 0.497 J

Step-by-step explanation:

Spring energy
Spring energy is stored in a compressible or stretchable object like a spring or rubber band.

The work done on a spring in changing its original length is converted to its elastic potential energy or spring energy.

Given -

The spring constant is k = 20N/m,

The initial stretch is x1 = 0.16m,

The additional stretch is x2 = 0.063m,

W = 0.5k (x1 + x2)^2 - 0.5k x1^2
W = 0.5 (20 N/m) (0.16m + 0.063m)^2
W = 0.497 J

User Kemal Tezer Dilsiz
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