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

Work Done by a Spring Force — Question 4

Section 7-6 · Physics pages 958–959

Question 4 Two work measurements can test the Hooke-law model Assume ideal massless Hookean springs with fixed supports unless stated otherwise. A fixed spring does −3 J when stretched from 0.10 to 0.20 m. A second measurement reports −6 J from 0.20 to 0.30 m. Treat these values as exact. x (m) .10 .20 .30 W1 = −3 J W2 = −6 J Two measured interval works must be consistent with one stiffness. (a) Infer k from the first measurement. (b) Predict the second work under the ideal model. (c) Can one Hookean stiffness fit both measurements? Explain what the discrepancy establishes.
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Question 4 - Solution Two work measurements can test the Hooke-law model x (m) .10 .20 .30 W1 = −3 J W2 = −6 J (a) Solve the first squared-extension relation. The governing relation is k = − 2Ws x2 f −x2 i −3 = 1 2k(.01 −.04) gives k = 200 N/m. (b) Test without refitting. W2,predicted = 100(.04 −.09) = −5 J. (c) Separate a failed model from a diagnosed cause. The reported −6 J would require k = 240 N/m on the second interval. No single ideal stiffness fits both exact results. This does not identify whether nonlinearity, calibration error, or an omitted force caused the failure.

Question 4

Two work measurements can test the Hooke-law model

Assume ideal massless Hookean springs with fixed supports unless stated otherwise. A fixed spring does −3-3 J when stretched from 0.10 to 0.20 m. A second measurement reports −6-6 J from 0.20 to 0.30 m. Treat these values as exact.

Diagram shown in the original SVG worksheet.

  1. Infer k from the first measurement.

  2. Predict the second work under the ideal model.

  3. Can one Hookean stiffness fit both measurements? Explain what the discrepancy establishes.

Question 4 - Solution

Two work measurements can test the Hooke-law model

Diagram shown in the original SVG worksheet.

(a) Solve the first squared-extension relation.

The governing relation is k=−2Wsxf2−xi2\boxed{k=-\frac{2W_s}{x_f^2-x_i^2}}

−3=12k(.01−.04)-3=\tfrac12k(.01-.04) gives k=200N/m.k=200\ \mathrm{N/m}.

(b) Test without refitting.

W2,predicted=100(.04−.09)=−5J.W_{2,\mathrm{predicted}}=100(.04-.09)=-5\ \mathrm J.

(c) Separate a failed model from a diagnosed cause.

The reported −6J-6\ \mathrm J would require k=240N/mk=240\ \mathrm{N/m} on the second interval. No single ideal stiffness fits both exact results. This does not identify whether nonlinearity, calibration error, or an omitted force caused the failure.

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