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

Work and Potential Energy — Question 6

Section 8-1 · Physics pages 1022–1023

Question 6 Two interaction energies change at once A 1 kg object moves upward by 0.50 m while its attached ideal spring changes extension from 0.30 to 0.10 m. The spring has k = 100 N/m; use g = 10 m/s2. Other agents may control the path. initial final ∆y = .50 m xs = .30 m xs = .10 m Height rises while the spring extension decreases. (a) Find ∆Ug and ∆Us. (b) Find the combined work of gravity and the spring. (c) If the object begins and ends at rest, find work by all other agents together.
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Question 6 - Solution Two interaction energies change at once initial final ∆y = .50 m xs = .30 m xs = .10 m (a) Evaluate the two configurations separately. Evaluate the two independent configuration changes: ∆Ug = 1(10)(.50) = 5 J , ∆Us = 1 2(100)(.102 −.302) = −4 J . (b) Add potential changes before reversing the sign. Add the potential changes before reversing the sign: Wg + Ws = −(∆Ug + ∆Us) = −(5 −4) = −1 J . The spring supplies 4 J while gravity absorbs 5 J. (c) Use the endpoint kinetic energies. The endpoint speeds vanish, so 0 = ∆K = Wg + Ws + Wother ⇒Wother = +1 J . Zero net work permits nonzero work by each separate interaction.

Question 6

Two interaction energies change at once

A 1 kg object moves upward by 0.50 m while its attached ideal spring changes extension from 0.30 to 0.10 m. The spring has k=100k=100 N/m; use g=10g=10 m/s2^2. Other agents may control the path.

Diagram shown in the original SVG worksheet.

  1. Find ΔUg\Delta U_g and ΔUs\Delta U_s.

  2. Find the combined work of gravity and the spring.

  3. If the object begins and ends at rest, find work by all other agents together.

Question 6 - Solution

Two interaction energies change at once

Diagram shown in the original SVG worksheet.

(a) Evaluate the two configurations separately.

Evaluate the two independent configuration changes: ΔUg=1(10)(.50)=5J,\boxed{\Delta U_g=1(10)(.50)=5\ \mathrm J}, ΔUs=12(100)(.102−.302)=−4J.\boxed{\Delta U_s=\tfrac12(100)(.10^2-.30^2)=-4\ \mathrm J}.

(b) Add potential changes before reversing the sign.

Add the potential changes before reversing the sign: Wg+Ws=−(ΔUg+ΔUs)=−(5−4)=−1J.\boxed{W_g+W_s=-(\Delta U_g+\Delta U_s)=-(5-4)=-1\ \mathrm J}. The spring supplies 4 J while gravity absorbs 5 J.

(c) Use the endpoint kinetic energies.

The endpoint speeds vanish, so 0=ΔK=Wg+Ws+Wother⇒Wother=+1J.0=\Delta K=W_g+W_s+W_{\mathrm{other}}\Rightarrow\boxed{W_{\mathrm{other}}=+1\ \mathrm J}. Zero net work permits nonzero work by each separate interaction.

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