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

Path Independence of Conservative Forces — Question 1

Section 8-2 · Physics pages 1032–1033

Question 1 A winding route through a uniform gravitational field A 2 kg parcel moves from height 1 m to 4 m. Route I rises directly; route II first rises to 7 m and then descends to 4 m. Use g = 10 m/s2. 1 m 4 m 7 m The vertical legs are separated horizontally only to make the two routes readable. (a) Calculate gravity work on both routes. (b) Find gravity work around the closed trip outward on I and back on II. (c) If a 1 N resistance opposes motion along each route, does its closed-trip work vanish?
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Question 1 - Solution A winding route through a uniform gravitational field 1 m 4 m 7 m (a) Sum signed vertical displacements. Gravity work depends on the signed height change: WI = −mg(4 −1) = −60 J , WII = −20(7 −1) −20(4 −7) = −120 + 60 = −60 J . (b) Reverse the endpoint change on the return. The return reverses the endpoint difference: WII,return = +60 J ⇒ I⃗ Fg · d⃗r = −60 + 60 = 0 . (c) Track resistance with distance. Resistance depends on total distance: sI = 3 m, sII = 6 + 3 = 9 m, Wr,loop = −1(3 + 9) = −12 J . The nonzero resistive loss does not contradict gravity’s separate path independence.

Question 1

A winding route through a uniform gravitational field

A 2 kg parcel moves from height 1 m to 4 m. Route I rises directly; route II first rises to 7 m and then descends to 4 m. Use g=10g=10 m/s2^2.

Diagram shown in the original SVG worksheet.

  1. Calculate gravity work on both routes.

  2. Find gravity work around the closed trip outward on I and back on II.

  3. If a 1 N resistance opposes motion along each route, does its closed-trip work vanish?

Question 1 - Solution

A winding route through a uniform gravitational field

Diagram shown in the original SVG worksheet.

(a) Sum signed vertical displacements.

Gravity work depends on the signed height change: WI=−mg(4−1)=−60J,\boxed{W_I=-mg(4-1)=-60\ \mathrm J}, WII=−20(7−1)−20(4−7)=−120+60=−60J.\boxed{W_{II}=-20(7-1)-20(4-7)=-120+60=-60\ \mathrm J}.

(b) Reverse the endpoint change on the return.

The return reverses the endpoint difference: WII,return=+60J⇒∮F→g⋅dr→=−60+60=0.W_{II,\mathrm{return}}=+60\ \mathrm J\Rightarrow\boxed{\oint\vec F_g\cdot d\vec r=-60+60=0}.

(c) Track resistance with distance.

Resistance depends on total distance: sI=3m,sII=6+3=9m,Wr,loop=−1(3+9)=−12J.s_I=3\ \mathrm m,\quad s_{II}=6+3=9\ \mathrm m,\qquad\boxed{W_{r,\mathrm{loop}}=-1(3+9)=-12\ \mathrm J}. The nonzero resistive loss does not contradict gravity’s separate path independence.

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