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

Work and Potential Energy — Question 8

Section 8-1 · Physics pages 1026–1027

Question 8 What a potential value can reveal about force Two one-dimensional interactions have U1(x) = 5 + 2x J and U2(x) = 5 −3x J, with x inserted in meters. For a small displacement, use Fx = −∆U/∆x; both functions are linear. x U U1 U2 5 J The two straight potential functions share an intercept but have opposite slopes. (a) Compare the two potential values at x=0. (b) Find the constant force for each interaction. (c) Explain why the common potential value cannot determine either force.
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Question 8 - Solution What a potential value can reveal about force x U U1 U2 5 J (a) Evaluate at the same position. At the common coordinate, U1(0) = 5 J = U2(0) . (b) Use the change per unit displacement. The negative slopes determine the forces: F1 = −dU1 dx = −2 N , F2 = −dU2 dx = +3 N . (c) Separate the offset from the spatial variation. A reference shift changes the value but not the slope: d(U + C) dx = dU dx , F = −dU dx . Equal potentials at one point therefore impose no equality on force magnitude or direction.

Question 8

What a potential value can reveal about force

Two one-dimensional interactions have U1(x)=5+2xU_1(x)=5+2x J and U2(x)=5−3xU_2(x)=5-3x J, with x inserted in meters. For a small displacement, use Fx=−ΔU/ΔxF_x=-\Delta U/\Delta x; both functions are linear.

Diagram shown in the original SVG worksheet.

  1. Compare the two potential values at x=0.

  2. Find the constant force for each interaction.

  3. Explain why the common potential value cannot determine either force.

Question 8 - Solution

What a potential value can reveal about force

Diagram shown in the original SVG worksheet.

(a) Evaluate at the same position.

At the common coordinate, U1(0)=5J=U2(0).\boxed{U_1(0)=5\ \mathrm J=U_2(0)}.

(b) Use the change per unit displacement.

The negative slopes determine the forces: F1=−dU1dx=−2N,F2=−dU2dx=+3N.\boxed{F_1=-\frac{dU_1}{dx}=-2\ \mathrm N},\qquad\boxed{F_2=-\frac{dU_2}{dx}=+3\ \mathrm N}.

(c) Separate the offset from the spatial variation.

A reference shift changes the value but not the slope: d(U+C)dx=dUdx,F=−dUdx.\frac{d(U+C)}{dx}=\frac{dU}{dx},\qquad F=-\frac{dU}{dx}. Equal potentials at one point therefore impose no equality on force magnitude or direction.

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