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Physics / Chapter 8: Potential Energy and Conservation of Energy / Section 8-2

Path Independence of Conservative Forces — Question 10

Section 8-2 · Physics pages 1050–1051

Question 10 The same geometric route traversed at different speeds A resistive force is⃗Fd = −b⃗v with b = 2 N s/m. An agent moves an object straight from A to B over 6 m, once at 1 m/s and once at 3 m/s. A B v = 1 or 3 m/s⃗ Fd = −b⃗v distance: 6 m Drag opposes the prescribed motion, and its magnitude depends on speed. (a) Find drag work on each trip. (b) Can a position-only potential reproduce both results? (c) If the object returns over the same 6 m at 1 m/s after the faster trip, find total drag work around the geometric loop.
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Question 10 - Solution The same geometric route traversed at different speeds A B v = 1 or 3 m/s⃗ Fd = −b⃗v distance: 6 m (a) Use the constant force on each prescribed trip. At constant speed,⃗Fd · d⃗r = −bv ds: W1 = −2(1)(6) = −12 J , W3 = −2(3)(6) = −36 J . (b) Compare equal position endpoints. A position-only potential would predict one value, WAB = U(A) −U(B), for the same endpoints. The two unequal works disprove that description because the speed histories differ. (c) Add two negative contributions. The return at 1 m/s also has negative drag work: Wloop = −36 −12 = −48 J . Reversing velocity reverses the drag vector along with the displacement, leaving their dot product negative.

Question 10

The same geometric route traversed at different speeds

A resistive force is F→d=−bv→\vec F_d=-b\vec v with b=2b=2 N s/m. An agent moves an object straight from A to B over 6 m, once at 1 m/s and once at 3 m/s.

Diagram shown in the original SVG worksheet.

  1. Find drag work on each trip.

  2. Can a position-only potential reproduce both results?

  3. If the object returns over the same 6 m at 1 m/s after the faster trip, find total drag work around the geometric loop.

Question 10 - Solution

The same geometric route traversed at different speeds

Diagram shown in the original SVG worksheet.

(a) Use the constant force on each prescribed trip.

At constant speed, F→d⋅dr→=−bvds\vec F_d\cdot d\vec r=-bv\,ds: W1=−2(1)(6)=−12J,W3=−2(3)(6)=−36J.\boxed{W_1=-2(1)(6)=-12\ \mathrm J},\qquad\boxed{W_3=-2(3)(6)=-36\ \mathrm J}.

(b) Compare equal position endpoints.

A position-only potential would predict one value, WAB=U(A)−U(B),W_{AB}=U(A)-U(B), for the same endpoints. The two unequal works disprove that description because the speed histories differ.

(c) Add two negative contributions.

The return at 1 m/s also has negative drag work: Wloop=−36−12=−48J.\boxed{W_{\mathrm{loop}}=-36-12=-48\ \mathrm J}. Reversing velocity reverses the drag vector along with the displacement, leaving their dot product negative.

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