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Conservative Forces and Potential Energy

Path independence, potential energy functions, and mechanical energy conservation

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Conservative Forces and Potential Energy

Conservative Force Definition

A force is conservative if the work done is independent of path (depends only on endpoints).

Equivalently:

  1. Work around any closed path is zero: ∮F⃗⋅dr⃗=0\oint \vec{F} \cdot d\vec{r} = 0
  2. The force can be written as: F⃗=−∇U\vec{F} = -\nabla U (gradient of scalar potential)

Potential Energy

For a conservative force: W=−ΔU=−(Uf−Ui)W = -\Delta U = -(U_f - U_i)

Differential form: dW=F⃗⋅dr⃗=−dUdW = \vec{F} \cdot d\vec{r} = -dU

Finding U from F

In one dimension: Fx=−dUdxF_x = -\frac{dU}{dx}

U(x)=−∫Fx dx+CU(x) = -\int F_x \, dx + C

In three dimensions: F⃗=−∇U=−(∂U∂xi^+∂U∂yj^+∂U∂zk^)\vec{F} = -\nabla U = -\left(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k}\right)

Finding F from U

Fx=−∂U∂x,Fy=−∂U∂y,Fz=−∂U∂zF_x = -\frac{\partial U}{\partial x}, \quad F_y = -\frac{\partial U}{\partial y}, \quad F_z = -\frac{\partial U}{\partial z}

Common Potential Energies

Gravitational (near Earth)

Ug=mghU_g = mgh

Fy=−dUdy=−mgF_y = -\frac{dU}{dy} = -mg

Spring

Us=12kx2U_s = \frac{1}{2}kx^2

Fx=−dUdx=−kxF_x = -\frac{dU}{dx} = -kx

Universal Gravitation

Ug=−Gm1m2rU_g = -\frac{Gm_1m_2}{r}

Fr=−dUdr=−Gm1m2r2F_r = -\frac{dU}{dr} = -\frac{Gm_1m_2}{r^2}

(Choosing U=0U = 0 at r=∞r = \infty)

Electric Potential Energy

Ue=kq1q2rU_e = k\frac{q_1q_2}{r}

Fr=−dUdr=−kq1q2r2F_r = -\frac{dU}{dr} = -k\frac{q_1q_2}{r^2}

Conservation of Mechanical Energy

For conservative forces only: E=KE+U=constantE = KE + U = \text{constant}

12mv2+U(x)=E\frac{1}{2}mv^2 + U(x) = E

Taking time derivative: mvdvdt+dUdt=0mv\frac{dv}{dt} + \frac{dU}{dt} = 0

mvdvdt+dUdxdxdt=0mv\frac{dv}{dt} + \frac{dU}{dx}\frac{dx}{dt} = 0

ma+dUdxv=0ma + \frac{dU}{dx}v = 0

F=−dUdxF = -\frac{dU}{dx} (recovers F⃗=−∇U\vec{F} = -\nabla U)

Equilibrium Points

At equilibrium, F=0F = 0: dUdx=0\frac{dU}{dx} = 0

Stable equilibrium: d2Udx2>0\frac{d^2U}{dx^2} > 0 (local minimum of UU)

Unstable equilibrium: d2Udx2<0\frac{d^2U}{dx^2} < 0 (local maximum of UU)

Neutral equilibrium: d2Udx2=0\frac{d^2U}{dx^2} = 0 (UU is flat)

Example: Potential Energy Curve

U(x)=12kx2−16bx3U(x) = \frac{1}{2}kx^2 - \frac{1}{6}bx^3

Equilibrium points: dUdx=kx−12bx2=0\frac{dU}{dx} = kx - \frac{1}{2}bx^2 = 0

x=0 or x=2kbx = 0 \text{ or } x = \frac{2k}{b}

Stability: d2Udx2=k−bx\frac{d^2U}{dx^2} = k - bx

At x=0x = 0: d2Udx2=k>0\frac{d^2U}{dx^2} = k > 0 (stable)

At x=2kbx = \frac{2k}{b}: d2Udx2=k−2k=−k<0\frac{d^2U}{dx^2} = k - 2k = -k < 0 (unstable)

Energy Diagrams

Plot U(x)U(x) vs xx. For total energy EE:

  • Particle confined to regions where E≥U(x)E \geq U(x)
  • Turning points where E=U(x)E = U(x) (velocity = 0)
  • Kinetic energy: KE=E−U(x)KE = E - U(x)

Non-Conservative Forces

When non-conservative forces (friction, drag) are present:

Wnc=ΔKE+ΔU=ΔEW_{nc} = \Delta KE + \Delta U = \Delta E

Mechanical energy is not conserved; it decreases by work done against non-conservative forces.

📚 Practice Problems

1Problem 1medium

❓ Question:

A particle moves in one dimension with potential energy U(x) = 4x² - x⁴ J (where x is in meters). Find: (a) the force as a function of position, (b) the equilibrium positions, and (c) determine which equilibria are stable.

💡 Show Solution

Given: U(x)=4x2−x4 JU(x) = 4x^2 - x^4 \text{ J}

(a) Force as function of position:

F(x)=−dUdx=−(8x−4x3)F(x) = -\frac{dU}{dx} = -(8x - 4x^3)

F(x)=4x3−8x N\boxed{F(x) = 4x^3 - 8x \text{ N}}

(b) Equilibrium positions:

Set F(x) = 0: 4x3−8x=04x^3 - 8x = 0 4x(x2−2)=04x(x^2 - 2) = 0

x=0,x=±2=±1.41 m\boxed{x = 0, \quad x = \pm\sqrt{2} = \pm 1.41 \text{ m}}

(c) Stability:

Test using second derivative: d2Udx2=8−12x2\frac{d^2U}{dx^2} = 8 - 12x^2

At x = 0: d2Udx2∣x=0=8>0\frac{d^2U}{dx^2}\Big|_{x=0} = 8 > 0 → Stable minimum

At x = ±√2: d2Udx2∣x=±2=8−12(2)=−16<0\frac{d^2U}{dx^2}\Big|_{x=\pm\sqrt{2}} = 8 - 12(2) = -16 < 0 → Unstable maxima

Visual:

  • U(0) = 0 (local minimum, stable)
  • U(±√2) = 4(2) - 4 = 4 J (local maxima, unstable)

Particle oscillates around x = 0 if energy E < 4 J.

2Problem 2hard

❓ Question:

A 0.5 kg particle moves under force F⃗=(2xy)i^+(x2−3y2)j^\vec{F} = (2xy)\hat{i} + (x^2 - 3y^2)\hat{j} N. Determine: (a) if the force is conservative, (b) if so, find the potential energy function, and (c) if the particle moves from (0,0) to (2,1) m, find the work done.

💡 Show Solution

Given: F⃗=2xyi^+(x2−3y2)j^\vec{F} = 2xy\hat{i} + (x^2 - 3y^2)\hat{j}

(a) Is force conservative?

Test: ∇×F⃗=0\nabla \times \vec{F} = 0

∂Fy∂x=∂∂x(x2−3y2)=2x\frac{\partial F_y}{\partial x} = \frac{\partial}{\partial x}(x^2 - 3y^2) = 2x

∂Fx∂y=∂∂y(2xy)=2x\frac{\partial F_x}{\partial y} = \frac{\partial}{\partial y}(2xy) = 2x

Since ∂Fy∂x=∂Fx∂y\frac{\partial F_y}{\partial x} = \frac{\partial F_x}{\partial y}:

Force is CONSERVATIVE\boxed{\text{Force is CONSERVATIVE}}

(b) Potential energy function:

Fx=−∂U∂x=2xy  ⟹  ∂U∂x=−2xyF_x = -\frac{\partial U}{\partial x} = 2xy \implies \frac{\partial U}{\partial x} = -2xy

Integrating with respect to x: U=−x2y+f(y)U = -x^2 y + f(y)

Fy=−∂U∂y=x2−3y2F_y = -\frac{\partial U}{\partial y} = x^2 - 3y^2

−∂∂y(−x2y+f(y))=x2−3y2-\frac{\partial}{\partial y}(-x^2 y + f(y)) = x^2 - 3y^2

x2−f′(y)=x2−3y2x^2 - f'(y) = x^2 - 3y^2

f′(y)=3y2  ⟹  f(y)=y3+Cf'(y) = 3y^2 \implies f(y) = y^3 + C

U(x,y)=−x2y+y3 J\boxed{U(x,y) = -x^2 y + y^3 \text{ J}} (taking C = 0)

(c) Work from (0,0) to (2,1):

For conservative force, work is independent of path: W=−ΔU=U(0,0)−U(2,1)W = -\Delta U = U(0,0) - U(2,1)

U(0,0)=0U(0,0) = 0 U(2,1)=−(2)2(1)+(1)3=−4+1=−3 JU(2,1) = -(2)^2(1) + (1)^3 = -4 + 1 = -3 \text{ J}

W=0−(−3)W = 0 - (-3)

W=3 J\boxed{W = 3 \text{ J}}

3Problem 3medium

❓ Question:

A spring with spring constant k = 200 N/m is compressed by x = 0.3 m from its equilibrium position. A 2.0 kg block is placed against it and released. Find: (a) the elastic potential energy stored, (b) the maximum speed of the block, and (c) the speed when the spring has returned halfway to equilibrium.

💡 Show Solution

Given:

  • k = 200 N/m
  • x₀ = 0.3 m (compressed)
  • m = 2.0 kg

(a) Elastic potential energy:

Us=12kx02=12(200)(0.3)2U_s = \frac{1}{2}kx_0^2 = \frac{1}{2}(200)(0.3)^2

Us=100(0.09)U_s = 100(0.09)

Us=9.0 J\boxed{U_s = 9.0 \text{ J}}

(b) Maximum speed:

At maximum speed, all elastic PE converts to KE: 12kx02=12mvmax2\frac{1}{2}kx_0^2 = \frac{1}{2}mv_{max}^2

vmax=kmx0=2002.0(0.3)v_{max} = \sqrt{\frac{k}{m}}x_0 = \sqrt{\frac{200}{2.0}}(0.3)

vmax=100(0.3)=10(0.3)v_{max} = \sqrt{100}(0.3) = 10(0.3)

vmax=3.0 m/s\boxed{v_{max} = 3.0 \text{ m/s}}

(Occurs when spring passes through equilibrium)

(c) Speed at halfway point:

At x = 0.15 m (halfway):

Energy conservation: 12kx02=12kx2+12mv2\frac{1}{2}kx_0^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2

9.0=12(200)(0.15)2+12(2.0)v29.0 = \frac{1}{2}(200)(0.15)^2 + \frac{1}{2}(2.0)v^2

9.0=2.25+v29.0 = 2.25 + v^2

v2=6.75v^2 = 6.75

v=2.6 m/s\boxed{v = 2.6 \text{ m/s}}

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Path independence, potential energy functions, and mechanical energy conservation
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