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Physics / Chapter 7: Kinetic Energy and Work / Section 7-6

Work Done by a Spring Force — Question 5

Section 7-6 · Physics pages 960–961

Question 5 External stretching work includes the change in kinetic energy Assume ideal massless Hookean springs with fixed supports unless stated otherwise. A 2 kg slider starts at rest with its horizontal spring relaxed. An agent moves it to extension 0.20 m, where its speed is 3 m/s. Here k = 100 N/m; there is no friction. m x The agent stretches the spring while also accelerating the slider. (a) Find spring work and the kinetic-energy change. (b) Find the work of the agent. (c) How would the agent work change if the same endpoint were reached quasistatically?
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Question 5 - Solution External stretching work includes the change in kinetic energy m x (a) Evaluate both known contributions. The governing relation is Wa + Ws = ∆K =⇒Wa = ∆K + 1 2k(x2 f −x2 i ) Ws = −1 2(100)(.20)2 = −2 J, and ∆K = 1 2(2)(3)2 = 9 J. (b) Use the total work theorem. Wa + Ws = ∆K, so Wa = 11 J. The applied force history is not determined by these endpoint data. (c) Specify the endpoint speed condition. In an ideal quasistatic move, ∆K ≈0, so Wa ≈2 J. Opposite agent and spring works require negligible kinetic-energy change, not merely the presence of a spring.

Question 5

External stretching work includes the change in kinetic energy

Assume ideal massless Hookean springs with fixed supports unless stated otherwise. A 2 kg slider starts at rest with its horizontal spring relaxed. An agent moves it to extension 0.20 m, where its speed is 3 m/s. Here k=100k=100 N/m; there is no friction.

Diagram shown in the original SVG worksheet.

  1. Find spring work and the kinetic-energy change.

  2. Find the work of the agent.

  3. How would the agent work change if the same endpoint were reached quasistatically?

Question 5 - Solution

External stretching work includes the change in kinetic energy

Diagram shown in the original SVG worksheet.

(a) Evaluate both known contributions.

The governing relation is Wa+Ws=ΔK⇒Wa=ΔK+12k(xf2−xi2)\boxed{W_a+W_s=\Delta K\Longrightarrow W_a=\Delta K+\tfrac12k(x_f^2-x_i^2)}

Ws=−12(100)(.20)2=−2J,W_s=-\tfrac12(100)(.20)^2=-2\ \mathrm J, and ΔK=12(2)(3)2=9J.\Delta K=\tfrac12(2)(3)^2=9\ \mathrm J.

(b) Use the total work theorem.

Wa+Ws=ΔK,W_a+W_s=\Delta K, so Wa=11J.W_a=11\ \mathrm J. The applied force history is not determined by these endpoint data.

(c) Specify the endpoint speed condition.

In an ideal quasistatic move, ΔK≈0\Delta K\approx0, so Wa≈2JW_a\approx2\ \mathrm J. Opposite agent and spring works require negligible kinetic-energy change, not merely the presence of a spring.

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