Determining Potential Energy Values — Question 3

Section 8-3 · Physics pages 1056–1057

Question 3 Joining two potential formulas at a force boundary For 0 ≤x ≤1 m, Fx = 3 N. For 1 ≤x ≤3 m, Fx = −2 N. Set U(0)=4 J; the force remains finite at the join. Fx x 3 N −2 N 1 m The force has a finite step; the potential must retain one common reference. (a) Find U(x) on the first interval. (b) Find a continuous U(x) on the second interval. (c) Find U(3) and explain why setting an independent zero at x=1 would be inconsistent.
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Question 3 - Solution Joining two potential formulas at a force boundary Fx x 3 N −2 N 1 m (a) Integrate from the reference. On the first interval, U(x) = U(0) − ∫x 0 3 ds = 4 −3x , 0 ≤x ≤1. The joining value is U(1) = 1 J. (b) Carry the joining value into the next interval. Carry that value into the second integral: U(x) = 1 − ∫x 1 (−2)ds = 1 + 2(x −1) = 2x −1 , 1 ≤x ≤3. (c) Preserve the physical continuity. At the final point, U(3) = 2(3) −1 = 5 J , U(1−) = U(1+) = 1 J. A bounded force contributes vanishing work over an infinitesimal interval. Resetting the zero at the join would introduce an unphysical jump and corrupt cross-boundary work.

Question 3

Joining two potential formulas at a force boundary

For 0≤x≤10\le x\le1 m, Fx=3F_x=3 N. For 1≤x≤31\le x\le3 m, Fx=−2F_x=-2 N. Set U(0)=4 J; the force remains finite at the join.

Diagram shown in the original SVG worksheet.

  1. Find U(x) on the first interval.

  2. Find a continuous U(x) on the second interval.

  3. Find U(3) and explain why setting an independent zero at x=1 would be inconsistent.

Question 3 - Solution

Joining two potential formulas at a force boundary

Diagram shown in the original SVG worksheet.

(a) Integrate from the reference.

On the first interval, U(x)=U(0)−∫0x3ds=4−3x,0≤x≤1.\boxed{U(x)=U(0)-\int_0^x3\,ds=4-3x},\quad0\leq x\leq1. The joining value is U(1)=1U(1)=1 J.

(b) Carry the joining value into the next interval.

Carry that value into the second integral: U(x)=1−∫1x(−2)ds=1+2(x−1)=2x−1,1≤x≤3.\boxed{U(x)=1-\int_1^x(-2)ds=1+2(x-1)=2x-1},\quad1\leq x\leq3.

(c) Preserve the physical continuity.

At the final point, U(3)=2(3)−1=5J,U(1−)=U(1+)=1J.\boxed{U(3)=2(3)-1=5\ \mathrm J},\qquad U(1^-)=U(1^+)=1\ \mathrm J. A bounded force contributes vanishing work over an infinitesimal interval. Resetting the zero at the join would introduce an unphysical jump and corrupt cross-boundary work.