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A spring with spring constant of 22 N/m is stretched 0.23 m from its equilibrium position. How much work must be done to stretch it

an additional 0.11 m? Answer in units of J.

User Leonidas
by
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1 Answer

3 votes
Answer:

0.1331 J

Step by step explanation:

The work required to stretch a spring is given by the formula:

W = (1/2)kx^2

where W is the work done, k is the spring constant, and x is the displacement from the equilibrium position.

The initial displacement of the spring is 0.23 m, and the final displacement is 0.23 m + 0.11 m = 0.34 m. Therefore, the displacement of the spring is Δx = 0.34 m - 0.23 m = 0.11 m.

The work required to stretch the spring an additional 0.11 m is:

W = (1/2)k(Δx)^2
= (1/2)(22 N/m)(0.11 m)^2
= 0.1331 J

Therefore, the work required to stretch the spring an additional 0.11 m is 0.1331 J.
User Mihaela Romanca
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7.7k points