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Physics / Chapter 8: Potential Energy and Conservation of Energy / Section 8-3

Determining Potential Energy Values — Question 5

Section 8-3 · Physics pages 1060–1061

Question 5 A shifted spring with a laboratory reference A spring of k = 100 N/m is relaxed when a slider is at q=0.30 m. Define U=0 at the laboratory coordinate q=0. q U q = .30 m U(q) The energy minimum occurs at the relaxed coordinate; the horizontal axis represents q. (a) Find U(q). (b) Find U at the relaxed position and at q=0.50 m. (c) Explain how an ideal spring can have negative U under this convention.
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Question 5 - Solution A shifted spring with a laboratory reference q U q = .30 m U(q) (a) Integrate in the deformation coordinate. The deformation is x = q −.30, so U(q) = 1 2k(q −.30)2 + C. The reference gives 0 = 50(.30)2 + C, hence U(q) = 50(q −.30)2 −4.5 J . (b) Evaluate relative to the assigned zero. Substitution yields U(.30) = −4.5 J , U(.50) = 50(.20)2 −4.5 = −2.5 J . (c) Identify what the negative sign means. The physically nonnegative deformation energy relative to the minimum is U(q) −Umin = 50(q −.30)2 ≥0. Negative assigned values mean energy below the chosen compressed reference, not negative stiffness or negative kinetic energy.

Question 5

A shifted spring with a laboratory reference

A spring of k=100k=100 N/m is relaxed when a slider is at q=0.30 m. Define U=0 at the laboratory coordinate q=0.

Diagram shown in the original SVG worksheet.

  1. Find U(q).

  2. Find U at the relaxed position and at q=0.50 m.

  3. Explain how an ideal spring can have negative U under this convention.

Question 5 - Solution

A shifted spring with a laboratory reference

Diagram shown in the original SVG worksheet.

(a) Integrate in the deformation coordinate.

The deformation is x=q−.30x=q-.30, so U(q)=12k(q−.30)2+C.U(q)=\tfrac12k(q-.30)^2+C. The reference gives 0=50(.30)2+C0=50(.30)^2+C, hence U(q)=50(q−.30)2−4.5J.\boxed{U(q)=50(q-.30)^2-4.5\ \mathrm J}.

(b) Evaluate relative to the assigned zero.

Substitution yields U(.30)=−4.5J,U(.50)=50(.20)2−4.5=−2.5J.\boxed{U(.30)=-4.5\ \mathrm J},\qquad\boxed{U(.50)=50(.20)^2-4.5=-2.5\ \mathrm J}.

(c) Identify what the negative sign means.

The physically nonnegative deformation energy relative to the minimum is U(q)−Umin=50(q−.30)2≥0.U(q)-U_{\min}=50(q-.30)^2\geq0. Negative assigned values mean energy below the chosen compressed reference, not negative stiffness or negative kinetic energy.

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