Determining Potential Energy Values — Question 6

Section 8-3 · Physics pages 1062–1063

Question 6 Recovering a quadratic potential from state measurements A potential is known to be U(x) = ax2 + bx + c. Measurements give U(0)=3 J, U(1 m)=4 J and U(2 m)=9 J. x U 3 4 9 1 2 Three measurements determine the coefficients only under the stipulated quadratic model. (a) Find a, b and c with units. (b) Find the associated force at x=1 m. (c) Predict work from x=0 to x=3 m and state which modeling assumption permits the prediction.
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Question 6 - Solution Recovering a quadratic potential from state measurements x U 3 4 9 1 2 (a) Solve the three state equations. Substitute the three measurements, using SI numerical coordinates: c = 3, a + b = 1, 4a + 2b = 6. Subtract twice the second relation from the third: a = 2 J/m2, b = −1 J/m, c = 3 J . (b) Differentiate the inferred function. The force is the negative derivative: Fx = −d dx(2x2 −x + 3) = 1 −4x ⇒Fx(1) = −3 N . (c) Extrapolate only under the stipulated form. Extrapolate the stipulated quadratic: U(3) = 2(9) −3 + 3 = 18 J, W0→3 = 3 −18 = −15 J . Three measurements alone would not justify this extrapolation without the assumed functional form.

Question 6

Recovering a quadratic potential from state measurements

A potential is known to be U(x)=ax2+bx+cU(x)=ax^2+bx+c. Measurements give U(0)=3 J, U(1 m)=4 J and U(2 m)=9 J.

Diagram shown in the original SVG worksheet.

  1. Find a, b and c with units.

  2. Find the associated force at x=1 m.

  3. Predict work from x=0 to x=3 m and state which modeling assumption permits the prediction.

Question 6 - Solution

Recovering a quadratic potential from state measurements

Diagram shown in the original SVG worksheet.

(a) Solve the three state equations.

Substitute the three measurements, using SI numerical coordinates: c=3,a+b=1,4a+2b=6.c=3,\qquad a+b=1,\qquad4a+2b=6. Subtract twice the second relation from the third: a=2J/m2,b=−1J/m,c=3J.\boxed{a=2\ \mathrm{J/m^2},\quad b=-1\ \mathrm{J/m},\quad c=3\ \mathrm J}.

(b) Differentiate the inferred function.

The force is the negative derivative: Fx=−ddx(2x2−x+3)=1−4x⇒Fx(1)=−3N.F_x=-\frac{d}{dx}(2x^2-x+3)=1-4x\Rightarrow\boxed{F_x(1)=-3\ \mathrm N}.

(c) Extrapolate only under the stipulated form.

Extrapolate the stipulated quadratic: U(3)=2(9)−3+3=18J,W0→3=3−18=−15J.U(3)=2(9)-3+3=18\ \mathrm J,\qquad\boxed{W_{0\to3}=3-18=-15\ \mathrm J}. Three measurements alone would not justify this extrapolation without the assumed functional form.