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Physics / Chapter 8: Potential Energy and Conservation of Energy / Section 8-1

Work and Potential Energy — Question 5

Section 8-1 · Physics pages 1020–1021

Question 5 A return to the same deformation An ideal spring of stiffness 80 N/m is slowly stretched from 0 to 0.30 m, then returned slowly to 0. Its support is fixed; kinetic-energy changes are negligible. slider An ideal controlled cycle stretches the spring and returns it to its relaxed configuration. (a) Find the spring potential-energy change and spring work on the outward leg. (b) Find spring work over the full cycle. (c) Find applied-agent work on each leg and explain why zero net cycle work does not mean no exchange occurred.
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Question 5 - Solution A return to the same deformation slider (a) Integrate the restoring force or use its potential. The stored spring energy rises by ∆Us = 1 2k(x2 f −x2 i ) = 1 2(80)(.30)2 = 3.6 J , Ws = −3.6 J . (b) Return to the same configuration. Returning to the original extension reverses the potential change: Ws,return = +3.6 J, Ws,cycle = −3.6 + 3.6 = 0 . (c) Follow the reversible exchanges. Quasistatic motion makes ∆K ≃0, so Wa = −Ws ⇒Wa,out = +3.6 J, Wa,back = −3.6 J . Energy is supplied and then recovered. Zero signed cycle work does not mean each transfer vanishes.

Question 5

A return to the same deformation

An ideal spring of stiffness 80 N/m is slowly stretched from 0 to 0.30 m, then returned slowly to 0. Its support is fixed; kinetic-energy changes are negligible.

Diagram shown in the original SVG worksheet.

  1. Find the spring potential-energy change and spring work on the outward leg.

  2. Find spring work over the full cycle.

  3. Find applied-agent work on each leg and explain why zero net cycle work does not mean no exchange occurred.

Question 5 - Solution

A return to the same deformation

Diagram shown in the original SVG worksheet.

(a) Integrate the restoring force or use its potential.

The stored spring energy rises by ΔUs=12k(xf2−xi2)=12(80)(.30)2=3.6J,Ws=−3.6J.\boxed{\Delta U_s=\tfrac12k(x_f^2-x_i^2)=\tfrac12(80)(.30)^2=3.6\ \mathrm J},\qquad\boxed{W_s=-3.6\ \mathrm J}.

(b) Return to the same configuration.

Returning to the original extension reverses the potential change: Ws,return=+3.6J,Ws,cycle=−3.6+3.6=0.W_{s,\mathrm{return}}=+3.6\ \mathrm J,\qquad\boxed{W_{s,\mathrm{cycle}}=-3.6+3.6=0}.

(c) Follow the reversible exchanges.

Quasistatic motion makes ΔK≃0\Delta K\simeq0, so Wa=−Ws⇒Wa,out=+3.6J,Wa,back=−3.6J.W_a=-W_s\Rightarrow\boxed{W_{a,\mathrm{out}}=+3.6\ \mathrm J,\quad W_{a,\mathrm{back}}=-3.6\ \mathrm J}. Energy is supplied and then recovered. Zero signed cycle work does not mean each transfer vanishes.

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