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

Work and Potential Energy — Question 4

Section 8-1 · Physics pages 1018–1019

Question 4 Negative potential values and positive kinetic energy A conservative interaction has UA = −35 J and UB = −20 J. A 2 kg particle at A has speed 5 m/s. No other force does work on it. A UA = −35 J B UB = −20 J The final potential is 15 J higher despite both assigned values being negative. (a) Find work by the interaction from A to B. (b) Find the kinetic energy and speed at B, if the path is accessible. (c) Do these endpoint data alone guarantee that the particle can reach B?
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Question 4 - Solution Negative potential values and positive kinetic energy A UA = −35 J B UB = −20 J (a) Subtract signed values. Subtract signed potential values before changing the sign: ∆U = −20 −(−35) = 15 J ⇒WAB = −15 J . (b) Apply the work theorem. The endpoint work–energy balance gives KA = 1 2(2)(5)2 = 25 J, KB = 25 −15 = 10 J . vB = √ 2KB/m = √ 10 = 3.162 m/s . (c) Retain the intermediate constraint. The conserved energy is E = KA + UA = 25 −35 = −10 J, K(s) = E −U(s) ≥0. An intermediate point with U(s) > −10 J blocks passage. The speed at B is conditional on an accessible route, which the endpoint data alone do not establish.

Question 4

Negative potential values and positive kinetic energy

A conservative interaction has UA=−35U_A=-35 J and UB=−20U_B=-20 J. A 2 kg particle at A has speed 5 m/s. No other force does work on it.

Diagram shown in the original SVG worksheet.

  1. Find work by the interaction from A to B.

  2. Find the kinetic energy and speed at B, if the path is accessible.

  3. Do these endpoint data alone guarantee that the particle can reach B?

Question 4 - Solution

Negative potential values and positive kinetic energy

Diagram shown in the original SVG worksheet.

(a) Subtract signed values.

Subtract signed potential values before changing the sign: ΔU=−20−(−35)=15J⇒WAB=−15J.\Delta U=-20-(-35)=15\ \mathrm J\Rightarrow\boxed{W_{AB}=-15\ \mathrm J}.

(b) Apply the work theorem.

The endpoint work–energy balance gives KA=12(2)(5)2=25J,KB=25−15=10J.K_A=\tfrac12(2)(5)^2=25\ \mathrm J,\qquad\boxed{K_B=25-15=10\ \mathrm J}. vB=2KB/m=10=3.162m/s.\boxed{v_B=\sqrt{2K_B/m}=\sqrt{10}=3.162\ \mathrm{m/s}}.

(c) Retain the intermediate constraint.

The conserved energy is E=KA+UA=25−35=−10J,K(s)=E−U(s)≥0.E=K_A+U_A=25-35=-10\ \mathrm J,\qquad K(s)=E-U(s)\geq0. An intermediate point with U(s)>−10U(s)>-10 J blocks passage. The speed at B is conditional on an accessible route, which the endpoint data alone do not establish.

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