Worked examples — Conservative forces — path-independent work, potential energy defined

Before anything, one reminder of the two tools we lean on the whole way:

The scenario matrix

Every problem in this topic lives in one of these cells. The examples that follow are tagged with the cell they cover.

Example 1 — Gravity up and down (cells A, B)

Forecast: Guess the sign of gravity's work going up, and going down. Do they cancel?

1. Write $U$ at each height. Why this step? We derived $U_{\text{grav}}=mgy$ in the parent, so we just read off the two piggy-bank values. $U_{\text{floor}}=mg(0)=0,U_{\text{shelf}}=mg(3)=2\cdot 10\cdot 3=60\text{J}.$
2. Leg up (cell A): Why this step? Use $W=U_A-U_B$ with $A=$floor, $B=$shelf. $W_{\uparrow }=U_{\text{floor}}-U_{\text{shelf}}=0-60=-60\text{J}.$ Gravity does negative work — it fights the lift, so it drains 60 J into the piggy bank.
3. Leg down (cell B): Why this step? Now $A=$shelf, $B=$floor — the endpoints swap. $W_{\downarrow }=U_{\text{shelf}}-U_{\text{floor}}=60-0=+60\text{J}.$ Gravity now does positive work — the piggy bank pays out its 60 J.
4. Round trip: $W_{\uparrow }+W_{\downarrow }=-60+60=0.$

Verify: The round trip is a closed loop, and gravity is conservative, so $\oint \vec{F}\cdot d\vec{r}$ must be $0$. It is. ✔ Units: $\text{kg}\cdot {\text{m/s}}^2\cdot \text{m}=\text{N}\cdot \text{m}=\text{J}$. ✔

The figure below draws this see-saw: read the magenta arrow (up, $W=-60$ J, gravity fighting) and the orange arrow (down, $W=+60$ J, gravity paying back). Notice the two $U$-labels on the left — $0$ J at the floor, $60$ J at the shelf — and that the arrows exactly undo each other, which is why the loop is zero.

Example 2 — Same start, same end, different roads (cell C)

Forecast: More steps, more distance travelled — does gravity charge more?

1. Kill the horizontal legs. Why this step? Gravity is $\vec{F}=-mg\hat{y}$; on any horizontal move $\vec{F}\cdot d\vec{r}=0$, so those three "right" legs contribute nothing.
2. Add the vertical legs. Why this step? Each "up $1$ m" leg does $W=-mg(1)=-20\text{J}$, and there are three of them. $W_{\text{stairs}}=3\times (-20)=-60\text{J}.$
3. Compare to the straight lift from Ex 1: $W_{\uparrow }=-60\text{J}$. Identical.

Verify: Path-independence is the definition of conservative — same endpoints ($0\to 3$ m) must give the same $-mgh=-2\cdot 10\cdot 3=-60\text{J}$ no matter how many corners we turn. ✔

Example 3 — Spring stretched vs compressed (cell D)

Forecast: Compression uses a negative $x$. Does that make $U$ negative?

1. Apply $U_{\text{spring}}=\frac{1}{2}k{x}^2$ for the stretch. Why this step? It's the formula we derived by integrating $F=-kx$. $U_{+}=\frac{1}{2}(200)(0.10{)}^2=\frac{1}{2}(200)(0.01)=1\text{J}.$
2. Now compress ($x=-0.10$). Why this step? Plug the negative $x$ in — but it is squared. $U_{-}=\frac{1}{2}(200)(-0.10{)}^2=\frac{1}{2}(200)(0.01)=1\text{J}.$ Same energy! The $x^2$ makes $U$ a symmetric bowl about $x=0$ — energy is stored whether you pull or push.
3. Forces differ in direction. Why this step? Use $F=-kx$, which keeps the sign of $x$. $F_{+}=-200(0.10)=-20\text{N}(\text{pulls back, toward }0),F_{-}=-200(-0.10)=+20\text{N}(\text{pushes out, toward }0).$

Verify: In both cases the force points toward $x=0$ — the bottom of the $U$-bowl, i.e. down the potential slope, exactly as $F=-dU/dx$ demands. The energies match by symmetry $(-x{)}^2=x^2$. ✔ Units: $\text{N/m}\cdot {\text{m}}^2=\text{N}\cdot \text{m}=\text{J}$. ✔ See Spring / elastic potential energy.

The figure shows the parabolic $U$-bowl. The magenta dot (stretch $+0.10$) and the orange dot (compress $-0.10$) sit at the same height $U=1$ J — that is the $(-x{)}^2=x^2$ symmetry. The short arrows near the bottom point inward, showing the force always shoves the mass back toward $x=0$, the base of the bowl.

Example 4 — Move the zero, change the number (cell E)

Forecast: Two different sets of numbers for $U$. Will the physics (the drop) change?

1. Floor reference. Why this step? $U=mgy$ with $y$ measured from the floor. $U_{\text{table}}=2\cdot 10\cdot 1=20\text{J},U_{\text{shelf}}=2\cdot 10\cdot 3=60\text{J}.$
2. Table reference. Why this step? Now $y^{\prime }=y-1$; the table is $y^{\prime }=0$, the shelf is $y^{\prime }=2$. $U_{\text{table}}^{\prime }=2\cdot 10\cdot 0=0\text{J},U_{\text{shelf}}^{\prime }=2\cdot 10\cdot 2=40\text{J}.$ Every number dropped by $20\text{J}$ — switching reference added a constant to $U$ everywhere.
3. Compare the drop table→shelf. Why this step? Only $\Delta U$ is physical. $\Delta U=U_{\text{shelf}}-U_{\text{table}}=60-20=40\text{J}=40-0=U_{\text{shelf}}^{\prime }-U_{\text{table}}^{\prime }.$

Verify: Call the constant we added when switching reference $U_0$ (here $U_0=-20$ J, since every value dropped by $20$ J). Because $U_0$ is a constant, its derivative is zero, so $F=-dU/dx$ is unchanged; and it cancels in any difference, so $\Delta U$ is unchanged too — indeed we got $40\text{J}$ both ways. ✔ The lesson: always state your reference.

Example 5 — A force that changes with position (cell F)

Forecast: The force grows steeply with $x$. Can we just multiply force by distance? (No — which tool do we need?)

1. Pick the right tool. Why this step? $F$ varies with $x$, so $W=F\Delta x$ is illegal; the correct machinery is the integral $U(x)=-\int Fdx$ (this is $F=-dU/dx$ read backwards). This is exactly the Work done by a variable force situation.
2. Integrate. Why this step? Undo the derivative to recover $U$. $U(x)=-\int (-b{x}^3)dx=b\int x^{3}dx=\frac{b{x}^4}{4}+C.$ Choose $U(0)=0\Rightarrow C=0$, so $U(x)=\frac{b{x}^4}{4}=x^4$ (since $b/4=1$).
3. Evaluate the two endpoints. Why this step? Energy released = drop in $U=U_A-U_B$. $U(1)=\frac{4\cdot 1^4}{4}=1\text{J},U(0)=0.$ $W_{1\to 0}=U(1)-U(0)=1-0=+1\text{J}.$

Verify: Check the force by differentiating: $-\frac{dU}{dx}=-\frac{d}{dx}\left(\frac{b{x}^4}{4}\right)=-b{x}^3=F(x)$. ✔ The direct work integral agrees: $W={\int }_1^0(-b{x}^3)dx={\left[-\frac{b{x}^4}{4}\right]}_1^0=0-(-1)=+1\text{J}$. ✔ Units: ${\text{N/m}}^3\cdot {\text{m}}^4=\text{N}\cdot \text{m}=\text{J}$. ✔

Example 6 — Friction around a loop (cell G)

Forecast: It's a closed loop. For gravity that gave $0$ (Ex 1). Will friction?

1. Leg out. Why this step? Friction always opposes motion, so it points backward while the block moves forward: $W_{\text{out}}=-fd=-3\cdot 5=-15\text{J}.$
2. Leg back. Why this step? The block now moves the other way, so friction flips too and still opposes: $W_{\text{back}}=-fd=-15\text{J}.$ It cannot cancel — it's negative on every leg.
3. Total loop work. $W_{\text{loop}}=-15+(-15)=-30\text{J}\ne 0.$

Verify: For a conservative force the loop must give $0$ (Ex 1 did). Here $W_{\text{loop}}=-30\text{J}\ne 0$, so friction is non-conservative and no $U(x)$ can exist. The stolen $30$ J became heat. ✔ See Friction and dissipation.

The figure makes the "always negative" point visual. On the upper track the block goes out (motion →) but the magenta friction arrow points ←; on the lower track it comes back (motion ←) yet the orange friction arrow now points → — friction has flipped to oppose motion again. Both legs cost $-15$ J, so the loop total ($-30$ J, in violet) can never reach zero.

Example 7 — The force that does nothing (cell H)

Forecast: A big $6$ N force acting the whole way around — surely lots of work?

1. Look at the geometry. Why this step? Work is $\vec{F}\cdot d\vec{r}=|F||dr|\cos \theta$; we need the angle between tension and motion. The ball moves along the circle (tangent); tension points to the centre (radius). Tangent ⟂ radius, so $\theta =90^{\circ }$.
2. Evaluate the dot product. Why this step? $\cos 90^{\circ }=0$. $dW=T|dr|\cos 90^{\circ }=0\text{at every instant}.$
3. Integrate around the loop. $W=\oint dW=0.$

Verify: This is the parent's "normal force / tension" point: a force that is always perpendicular to motion does zero work — not because of its name but because $\vec{F}\cdot d\vec{r}=0$. So it never dissipates and never stores energy. ✔ (Compare Gradient and vector fields: a force perpendicular to displacement contributes nothing to the line integral.)

Example 8 — Real-world energy budget (cell I)

Forecast: Without friction she'd convert all $mgh$ to $KE$. How much does the thief cost her?

1. Energy the piggy bank releases. Why this step? Gravity is conservative; the released energy is the drop $U_{\text{top}}-U_{\text{bottom}}=mgh$. $mgh=30\cdot 10\cdot 2=600\text{J}.$
2. Subtract the thief. Why this step? Friction is non-conservative and takes $60\text{J}$ as heat, so the kinetic energy actually delivered is the remainder (this is the Work–Energy theorem + Conservation of mechanical energy with a loss term). $K{E}_{\text{bottom}}=600-60=540\text{J}.$
3. Solve for speed. Why this step? $KE=\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{2KE/m}$. $v=\sqrt{\frac{2\cdot 540}{30}}=\sqrt{36}=6\text{m/s}.$

Verify: Without friction: $v_{\max }=\sqrt{2gh}=\sqrt{40}\approx 6.32\text{m/s}$, so friction should give something slightly less — and $6<6.32$. ✔ Units: $\sqrt{\text{J/kg}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}$. ✔

Example 9 — Exam twist: is this force conservative? (cell J)

Forecast: It's just a push in one direction — feels harmless and conservative. Trust the loop, not the vibe.

1. Leg $A\to B$ (along $x$, $y=0$). Why this step? $\vec{F}=(y,0)=(0,0)$ here, so $\int \vec{F}\cdot d\vec{r}=0$.
2. Leg $B\to C$ (along $y$, motion is $\hat{y}$). Why this step? $\vec{F}=(y,0)$ has no $y$-component, so $\vec{F}\cdot d\vec{r}=0$. Contribution $0$.
3. Leg $C\to D$ (along $x$, backward, $y=1$). Why this step? Now $\vec{F}=(1,0)$, motion is $-\hat{x}$ over length $1$: $\int =(1)(-1)=-1\text{J}$.
4. Leg $D\to A$ (along $y$, downward). Why this step? Again no $y$-component of force, contribution $0$.
5. Add the loop. $\oint \vec{F}\cdot d\vec{r}=0+0+(-1)+0=-1\text{J}\ne 0.$

Conclusion: The loop work is $-1\text{J}\ne 0$, so this innocent-looking single-direction push is non-conservative — no potential energy $U$ exists for it.

Verify: Cross-check with the Gradient and vector fields curl test: for a conservative 2D field $\vec{F}=(F_x,F_y)$ we need $\partial F_x/\partial y=\partial F_y/\partial x$. Here $\partial (y)/\partial y=1$ but $\partial (0)/\partial x=0$, and $1\ne 0$, so the field fails the test — confirming it is non-conservative. ✔ Both methods (loop integral and curl) agree.

Active recall

Connections

• Parent topic (Hinglish)
• Work done by a variable force
• Work–Energy theorem
• Conservation of mechanical energy
• Gravitational potential energy
• Spring / elastic potential energy
• Friction and dissipation
• Gradient and vector fields

Case Map