Exercises — Conservative forces — path-independent work, potential energy defined

WHAT: Add the work leg by leg around the closed loop. WHY: Conservative $\iff$ $\oint \vec{F}\cdot d\vec{r}=0$. $W_{\text{loop}}=W_{A\to B}+W_{B\to A}=(+6)+(-6)=0\text{J}.$ Both paths $A\to B$ gave the same $+6\text{J}$ (path-independent) and the loop is zero. Yes, it is consistent with a conservative force.

Friction (c). Its work is path-dependent — it always opposes motion, so a longer path loses more energy and the loop work is $\ne 0$. Gravity and springs are path-independent, so each has a $U$. Answer: friction. See Friction and dissipation.

Floor reference: $U=mgy=(2.0)(9.8)(1.5)=29.4\text{J}$. Ceiling reference: measure height from the ceiling, so the shelf is at $y^{\prime }=1.5-3.0=-1.5\text{m}$: $U^{\prime }=mg{y}^{\prime }=(2.0)(9.8)(-1.5)=-29.4\text{J}.$ Drop to the floor:

• Floor ref: $\Delta U=U_{\text{floor}}-U_{\text{shelf}}=0-29.4=-29.4\text{J}$.
• Ceiling ref: floor is at $y^{\prime }=-3.0$, so $U_{\text{floor}}^{\prime }=(2.0)(9.8)(-3.0)=-58.8\text{J}$; $\Delta U=-58.8-(-29.4)=-29.4\text{J}$.

Same $\Delta U=-29.4\text{J}$ even though the absolute $U$ values differ. Only differences are physical. See Gravitational potential energy.

(a) $U=\frac{1}{2}k{x}^2=\frac{1}{2}(200)(0.10{)}^2=1.0\text{J}$. (b) $F=-\frac{dU}{dx}=-kx=-(200)(0.10)=-20\text{N}$. The minus means the force points back toward $x=0$ (restoring). Magnitude $20\text{N}$. See Spring / elastic potential energy.

For a constant force, $W=\vec{F}\cdot \Delta \vec{r}$ no matter the path (that's the whole point). Path 1: $\Delta \vec{r}=(2,1)$, so $W_1=(3)(2)+(4)(1)=6+4=10\text{J}$. Path 2, leg 1 $(0,0)\to (2,0)$: $\Delta \vec{r}=(2,0)$, $W=(3)(2)+(4)(0)=6\text{J}$. Path 2, leg 2 $(2,0)\to (2,1)$: $\Delta \vec{r}=(0,1)$, $W=(3)(0)+(4)(1)=4\text{J}$. $W_2=6+4=10\text{J}=W_1.$ ✔ Path-independent, so this constant force is conservative.

Friction: opposes motion on both legs, so it is negative each way: $W_f=-fd-fd=-(5.0)(3.0)-(5.0)(3.0)=-30\text{J}\ne 0.$ Loop work non-zero $\Rightarrow$ non-conservative, no $U$. Gravity on a horizontal loop: force is vertical, motion horizontal, $\vec{F}\cdot d\vec{r}=0$ everywhere, so $W_{\text{grav}}=0$. Gravity gives zero loop work (as it must for a conservative force).

(a) WHAT: Undo $F=-dU/dx$, i.e. $dU=-Fdx$. WHY: $U$ is the anti-derivative of $-F$ (see Work done by a variable force). $U(x)=-\int Fdx=-\int (-kx+a{x}^3)dx=\frac{1}{2}k{x}^2-\frac{1}{4}a{x}^4+C.$ With $U(0)=0$, $C=0$: $U(x)=\frac{1}{2}(8)x^2-\frac{1}{4}(2)x^4=4x^2-\frac{1}{2}{x}^4.$ (b) Force zero: $F=0\Rightarrow -8x+2x^3=0\Rightarrow 2x(x^2-4)=0\Rightarrow x=0,+2,-2$ (in metres).

WHY the second derivative: $F=-dU/dx$, so near an equilibrium the force is like a spring $F\approx -(U^{\prime \prime })\delta$. If $U^{\prime \prime }>0$ (valley) the force pushes you back → stable; if $U^{\prime \prime }<0$ (hill) it pushes you away → unstable. $U^{\prime }(x)=8x-2x^3,U^{\prime \prime }(x)=8-6x^2.$

• $x=0$: $U^{\prime \prime }=8>0$ → stable (bottom of the valley).
• $x=\pm 2$: $U^{\prime \prime }=8-6(4)=-16<0$ → unstable (the two hilltops).

Look at the figure: the curve dips at $x=0$ and crests at $x=\pm 2$ — a ball placed exactly on a crest rolls off. This connects to Gradient and vector fields (force = $-\nabla U$).

WHAT: Walk the four edges, adding $\vec{F}\cdot d\vec{r}$ on each.

• $O\to A$: along $x$, $y=0$, $d\vec{r}=(dx,0)$, $\vec{F}\cdot d\vec{r}=ydx=0$. → $0$.
• $A\to B$: along $y$, $x=1$, $\vec{F}\cdot d\vec{r}=xdy=1dy$, ${\int }_0^1=1$.
• $B\to C$: along $x$ from 1→0, $y=1$, $\vec{F}\cdot d\vec{r}=ydx=1dx$, ${\int }_1^0=-1$.
• $C\to O$: along $y$ from 1→0, $x=0$, $\vec{F}\cdot d\vec{r}=xdy=0$. → $0$. $\oint =0+1-1+0=0.$ Zero loop work → conservative. Find $U$: need $-\partial U/\partial x=F_x=y$ and $-\partial U/\partial y=F_y=x$. Integrate the first: $U=-xy+h(y)$. Then $-\partial U/\partial y=x-h^{\prime }(y)=x\Rightarrow h^{\prime }(y)=0$. So $U(x,y)=-xy+C.$ Check: $\vec{F}=-\nabla U=-(-y,-x)=(y,x)$ ✔.

WHY energy conservation: the incline is frictionless and both spring and gravity are conservative, so mechanical energy is conserved (see Conservation of mechanical energy). Spring energy released fully converts to gravitational PE at the highest point (speed $=0$ there): $\frac{1}{2}k{x}^2=mg\Delta h,\Delta h=s\sin 30^{\circ }.$ Left side: $\frac{1}{2}(400)(0.20{)}^2=\frac{1}{2}(400)(0.04)=8.0\text{J}$. Solve for $s$: $s=\frac{\frac{1}{2}k{x}^2}{mg\sin 30^{\circ }}=\frac{8.0}{(0.50)(9.8)(0.5)}=\frac{8.0}{2.45}\approx 3.27\text{m}.$ The block slides $\approx 3.27\text{m}$ up the slope. (If you want the total climb measured from the compressed position, add the $0.20\text{m}$ the spring pushed through the incline.)

Q: What single test decides if a force has a potential energy? ::: Is its work path-independent — equivalently, is $\oint \vec{F}\cdot d\vec{r}=0$? If yes, $U$ exists with $\vec{F}=-\nabla U$.