Conservation of Energy

Overview of Energy Conservation

• Chapter 8 focuses on the concepts of work and energy, specifically introducing potential energy and the law of conservation of energy.

• The central principle of this chapter is that total energy remains constant in any process, as supported by scientific experiment.

• Energy conservation provides a technical tool to solve problems where Newton's laws are difficult to apply due to unknown or unmeasurable forces.

• The scope of the study is primarily restricted to particles or rigid objects undergoing translational motion without internal or rotational motion.

8-1 Conservative and Nonconservative Forces

• Definition of Conservative Force: A force is conservative if the work done by that force on an object moving from one point to another depends only on the initial and final positions of the object and is independent of the particular path taken.

• A conservative force must be a function of position only; it cannot depend on time or velocity.

• Gravity as a Conservative Force:

  • The gravitational force near Earth's surface is $\mathbf{F} = m\mathbf{g}$.

  • Work done by gravity on an object falling vertically by height $h$ is $W_G = Fd = mgh$.

  • For an arbitrary two-dimensional path in the xy plane from $y_1$ to $y_2$:     $W_G = \int_1^2 \mathbf{F}_G \cdot d\mathbf{l} = \int_1^2 mg \cos(\theta) dl$

  • Using the relationship $\cos(\theta) = -\cos(\phi)$ and $dy = dl \cos(\phi)$, where $\phi = 180^\circ - \theta$:     $W_G = -\int_{y_1}^{y_2} mg dy = -mg(y_2 - y_1)$ (Equation 8-1).

  • Since the work depends only on the vertical height $h = y_1 - y_2$, gravity is confirmed as a conservative force.

• Closed Path Definition: A force is conservative if the net work done by the force on an object moving around any closed path is zero ($W_{total} = W + (-W) = 0$).

• Recoverability: The work done by conservative forces is recoverable. Positive work done on one segment is negated by equivalent negative work on the return segment.

• Nonconservative Forces:

  • Forces such as friction, air resistance, tension in a cord, motor propulsion, and pushes/pulls by people are nonconservative.

  • Work done by nonconservative forces depends on the distance and path taken. For example, pushing a crate along a curved path requires more work against friction than a straight path because the distance $d$ is greater and the pushing force $\mathbf{F}_p$ is always in the direction of motion.

• Table 8-1: Categorization of Forces

  • Conservative Forces: Gravitational, Elastic, Electric.

  • Nonconservative Forces: Friction, Air resistance, Tension in cord, Motor or rocket propulsion, Push or pull by a person.

8-2 Potential Energy

• Definition: Potential energy ($U$) is the energy associated with forces that depend on the position or configuration of an object relative to its surroundings.

• Potential energy is only defined for conservative forces.

• Gravitational Potential Energy:

  • An object raised to height $h$ has the capacity to do work (e.g., a brick falling to drive a stake).

  • External work done to lift an object without acceleration: $W_{ext} = F_{ext} \cdot d = mgh \cos(0^\circ) = mg(y_2 - y_1)$.

  • Work done by gravity during the lift: $W_G = \mathbf{F}_G \cdot \mathbf{d} = mgh \cos(180^\circ) = -mgh$.

  • Change in Potential Energy (General): $\Delta U = U_2 - U_1 = W_{ext} = -W_G = mg(y_2 - y_1)$ (Equation 8-2).

  • Gravitational Potential Energy Formula: $U_{grav} = mgy$ (Equation 8-3).

• Reference Levels: The choice of reference level ($y = 0$) is arbitrary and determined for convenience. The physically meaningful quantity is the change in potential energy ($\Delta U$), which remains constant regardless of the reference point.

• System Property: Potential energy belongs to a system (e.g., object-Earth system), not a single object.

Example 8-1: Roller Coaster Potential Energy

• Scenario: A $1000\,kg$ car moves from Point 1 ($y=0$) to Point 2 ($10\,m$) to Point 3 ($-15\,m$ relative to 1).

• Part (a): Energy relative to Pt 1:

  • $U_2 = mgy_2 = (1000\,kg)(9.8\,m/s^2)(10\,m) = 9.8 \times 10^4\,J$.

  • $U_3 = mgy_3 = (1000\,kg)(9.8\,m/s^2)(-15\,m) = -1.5 \times 10^5\,J$.

• Part (b): Change from Pt 2 to Pt 3:

  • $\Delta U = U_3 - U_2 = (-1.5 \times 10^5\,J) - (9.8 \times 10^4\,J) = -2.5 \times 10^5\,J$.

• Part (c): Reference at Pt 3 ($y_3 = 0$):

  • $y_1 = 15\,m$, $y_2 = 25\,m$.

  • $U_2 = (1000\,kg)(9.8\,m/s^2)(25\,m) = 2.5 \times 10^5\,J$.

  • $\Delta U = U_3 - U_2 = 0 - 2.5 \times 10^5\,J = -2.5 \times 10^5\,J$.

General and Elastic Potential Energy

• General Change in PE: $\Delta U = U_2 - U_1 = - \int_1^2 \mathbf{F} \cdot d\mathbf{l} = -W$ (Equation 8-4).

• Elastic Potential Energy:

  • Spring force (Hooke's Law): $F_s = -kx$.

  • Potential energy stored in a spring compressed or stretched by distance $x$:     $U(x) - U(0) = - \int_0^x (-kx) dx = \frac{1}{2}kx^2$.

  • Assuming $U(0) = 0$, then $U_{el}(x) = \frac{1}{2}kx^2$ (Equation 8-5).

• Force Related to Potential Energy (1-D):

  • Given $F(x)$, $U(x) = - \int F(x) dx + C$ (Equation 8-6).

  • Given $U(x)$, $F(x) = -\frac{dU(x)}{dx}$ (Equation 8-7).

• Example: Determine F from U:

  • If $U(x) = -\frac{ax}{b^2 + x^2}$, then $F(x) = -\frac{d}{dx}\left(-\frac{ax}{b^2 + x^2}\right) = \frac{a(b^2 - x^2)}{(b^2 + x^2)^2}$.

• Potential Energy in Three Dimensions:

  • Relationship using partial derivatives:     $F_x = -\frac{\partial U}{\partial x}$, $Fy = -\frac{\partial U}{\partial y}$, $F_z = -\frac{\partial U}{\partial z}$.     $\mathbf{F}(x, y, z) = -\mathbf{i} \frac{\partial U}{\partial x} - \mathbf{j} \frac{\partial U}{\partial y} - \mathbf{k} \frac{\partial U}{\partial z}$.

8-3 Mechanical Energy and Its Conservation

• Work-Energy Principle: $W_{net} = \Delta K$.

• For a conservative system: $W_{net} = -\Delta U_{total}$.

• Combining these: $\Delta K + \Delta U = 0$ (Equation 8-9a).

• Total Mechanical Energy (E): $E = K + U$.

• Principle of Conservation of Mechanical Energy: If only conservative forces do work, the total mechanical energy of a system remains constant:   $K_2 + U_2 = K_1 + U_1$ (Equation 8-10).   $E_2 = E_1 = \text{constant}$.

• For a single object system: $\frac{1}{2}mv^2 + U = \text{constant}$ (Equation 8-11a).

8-4 Problem Solving with Mechanical Energy

• In the absence of air resistance, a falling rock transforms $U_{grav}$ into $K$.

  • Equation: $\frac{1}{2}mv_1^2 + mgy_1 = \frac{1}{2}mv_2^2 + mgy_2$ (Equation 8-12).

• Example 8-2/8-3: Falling Rock:

  • Rock dropped from $y_1 = 3.0\,m$, find speed at $y_2 = 1.0\,m$.

  • $mgy_1 = \frac{1}{2}mv_2^2 + mgy_2$

  • $v_2 = \sqrt{2g(y_1 - y_2)} = \sqrt{2(9.8\,m/s^2)(3.0\,m - 1.0\,m)} = 6.3\,m/s$.

• Exercise B: Calculated speed just before impact ($y_2 = 0$) for the rock: $v = \sqrt{2(9.8)(3.0)} \approx 7.7\,m/s$.

• Roller Coasters and Normal Force: The normal force acts perpendicular to motion, thus doing zero work ($W = 0$). Mechanical energy is conserved despite the normal force Presence.

• Example 8-4: Roller Coaster Speed:

  • Starts at top ($y_1 = 40\,m$). (a) Speed at bottom ($y_2 = 0$): $v_2 = \sqrt{2(9.8)(40)} = 28\,m/s$.

  • (b) Height where speed is half ($14\,m/s$): $y_2 = y_1 - \frac{v_2^2}{2g} = 40\,m - \frac{(14\,m/s)^2}{2(9.8\,m/s^2)} = 30\,m$.

• Example 8-5: Water Slides:

  • Paul and Jeanne start from same height $h$. Both have the same speed at the bottom because $\frac{1}{2}mv^2 = mgh$.

  • Jeanne finishes first because she is at a lower elevation earlier, meaning she has higher speed throughout most of the path.

Sports and Combined Energy Examples

• Example 8-6: Pole Vault:

  • $K_{running} \rightarrow U_{el, pole} \rightarrow U_{grav, vaulter}$.

  • A $70\,-kg$ vaulter needs to clear $5.0\,m$ bar. CM starts at $0.90\,m$. Height change $\Delta y = 4.1\,m$.

  • $K_1 = mgy_2 = (70\,kg)(9.8\,m/s^2)(4.1\,m) = 2.8 \times 10^3\,J$.

  • Required speed: $v_1 = \sqrt{\frac{2(2800\,J)}{70\,kg}} = 8.9\,m/s \approx 9\,m/s$.

• Example 8-7: Toy Dart Gun:

  • $m = 0.100\,kg$, $k = 250\,N/m$, compression $x = 0.060\,m$.

  • $\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

  • $v = \sqrt{\frac{kx^2}{m}} = \sqrt{\frac{(250\,N/m)(0.060\,m)^2}{0.100\,kg}} = 3.0\,m/s$.

• Example 8-8: Two kinds of Potential Energy:

  • Ball $m = 2.60\,kg$ falls distance $h = 0.550\,m$ then compresses spring $Y = 0.150\,m$.

  • Total conservation from Point 1 (start) to Point 3 (max compression):     $mgh = -mgY + \frac{1}{2}kY^2$

  • $k = \frac{2mg(h + Y)}{Y^2} = \frac{2(2.60\,kg)(9.80\,m/s^2)(0.550\,m + 0.150\,m)}{(0.150\,m)^2} = 1590\,N/m$.

• Example 8-9: Simple Pendulum:

  • Bob starts at angle $\theta_0$, cord length $L$. Initial height $y_0 = L(1 - \cos(\theta_0))$.

  • (b) Speed vs position: $v = \sqrt{2gL(\cos(\theta) - \cos(\theta_0))}$.

  • (c) Max speed (at $y=0$): $v = \sqrt{2gL(1 - \cos(\theta_0))}$.

  • (d) Tension $F_T$ using $\sum F = ma_r$: $F_T - mg \cos(\theta) = m\frac{v^2}{L}$.

  • Resulting tension: $F_T = (3 \cos(\theta) - 2 \cos(\theta_0))mg$.

8-5 & 8-6 Law of Conservation of Energy and Dissipative Forces

• General Law of Conservation of Energy: Total energy is neither increased nor decreased in any process. Change in total energy is zero:   $\Delta K + \Delta U + [\text{change in all other forms of energy}] = 0$ (Equation 8-14).

• Dissipative Forces: Forces like friction that reduce mechanical energy by transforming it into thermal energy (internal energy).

• General Work-Energy Principle:   $\Delta K + \Delta U = W_{NC}$ (Equation 8-15a).   Where $W_{NC}$ is work done by nonconservative forces.

• For a roller coaster with friction: $\frac{1}{2}mv_1^2 + mgy_1 = \frac{1}{2}mv_2^2 + mgy_2 + F_{fr}l$ (Equation 8-15b).

• Problem Solving Strategy Summary:

  1. Draw situation.

  2. Define system.

  3. Identify unknown and initial/final positions.

  4. Choose reference frame ($y=0$, uncompressed spring at $x=0$).

  5. Determine if mechanical energy is conserved.

  6. Apply appropriate equation ($K_1 + U_1 = K_2 + U_2$ or include $W_{NC}$).

  7. Solve for unknown.

• Example 8-10: Friction on Roller Coaster:

  • $m = 1000\,kg$, $y_1 = 40\,m$, $y_2 = 25\,m$, path length $l = 400\,m$. Car stops at $y_2$.

  • Thermal energy generated: $F_{fr}l = mg(y_1 - y_2) = (1000)(9.8)(15) = 1.47 \times 10^5\,J$.

  • Average friction force: $F_{fr} = (1.47 \times 10^5\,J) / 400\,m = 370\,N$.

• Example 8-11: Friction with a Spring:

  • Block of mass $m$ at speed $v_0$ hits spring, compresses distance $X$.

  • Initial energy ($\frac{1}{2}mv_0^2$) = Final storage ($\frac{1}{2}kX^2$) + Thermal work ($\mu_k mgX$).

  • $\mu_k = \frac{v_0^2}{2gX} - \frac{kX}{2mg}$.

8-7 Universal Gravitational Potential Energy and Escape Velocity

• For large distances from Earth center (r > r_E), the force is $F = -G \frac{m M_E}{r^2}$.

• Work done moving from $r_1$ to $r_2$: $W = \int_1^2 \mathbf{F} \cdot d\mathbf{l} = G m M_E \left( \frac{1}{r_2} - \frac{1}{r_1} \right)$.

• Change in PE: $\Delta U = -W = -G m M_E \left( \frac{1}{r_2} - \frac{1}{r_1} \right)$ (Equation 8-16).

• Choosing reference level $U = 0$ at $r = \infty$:   $U(r) = - \frac{G m M_E}{r}$ (Equation 8-17).

• Example 8-13: Dropping a Package from Space:

  • Box $m$ dropped from rocket at $v_1 = 1800\,m/s$ at altitude $1600\,km$ ($r_1 = 7.98 \times 10^6\,m$).

  • Impact at surface ($r_2 = 6.38 \times 10^6\,m$).

  • $v_2 = \sqrt{v_1^2 - 2GM_E \left(\frac{1}{r_1} - \frac{1}{r_2}\right)} = 5320\,m/s$.

• Escape Velocity ($v_{esc}$): The minimum initial velocity to leave Earth ($r_2 = \infty$, $v_2 = 0$).

  • $\frac{1}{2}m v_{esc}^2 - \frac{G m M_E}{r_E} = 0$

  • $v_{esc} = \sqrt{\frac{2GM_E}{r_E}} = 1.12 \times 10^4\,m/s = 11.2\,km/s$ (Equation 8-19).

• Example 8-14: Earth vs. Moon:

  • (a) Ratio of escape velocities: $\frac{v_{esc,Earth}}{v_{esc,Moon}} = \sqrt{\frac{M_E r_M}{M_M r_E}} \approx 4.7$.

  • (b) Energy ratio: Launching from Earth requires $(4.7)^2 \approx 22$ times more energy than the Moon.

8-8 Power

• Power (P): The rate at which work is done or energy is transformed.

  • Average Power: $P = \frac{W}{t}$ (Equation 8-20a).

  • Instantaneous Power: $P = \frac{dW}{dt}$ or $P = \frac{dE}{dt}$.

• Power and Force: $P = \frac{\mathbf{F} \cdot d\mathbf{l}}{dt} = \mathbf{F} \cdot \mathbf{v}$ (Equation 8-21).

• Units:

  • SI: Watt ($1\,W = 1\,J/s$).

  • British: $1\,horsepower (hp) = 550\,ft \cdot lb/s = 746\,W$.

  • $1\,kW \approx 1.34\,hp$.

• Example 8-15: Stair-Climbing Jogger:

  • $m = 60\,kg$, height $h = 4.5\,m$, time $t = 4.0\,s$.

  • $P = \frac{mgh}{t} = \frac{(60)(9.8)(4.5)}{4.0} = 660\,W \approx 0.88\,hp$.

• Example 8-16: Power Needs of a Car:

  • Car $m = 1400\,kg$, retarding force $F_R = 700\,N$.

  • (a) Climbing $10^\circ$ hill at $80\,km/h$ ($22\,m/s$):

  • Total force $F = F_R + mg \sin(10^\circ) = 700 + (1400)(9.8)(0.174) = 3100\,N$.

  • $P = Fv = (3100)(22) = 68.0\,kW = 91\,hp$.

  • (b) Accelerating from $90$ to $110\,km/h$ in $6.0\,s$:

  • $a = (30.6 - 25.0) / 6.0 = 0.93\,m/s^2$.

  • $F = ma + F_R = (1400)(0.93) + 700 = 2000\,N$.

  • Max Power required: $P = (2000\,N)(30.6\,m/s) = 61.2\,kW = 82\,hp$.

• Efficiency (e): Ratio of useful power output to power input ($e = P_{out} / P_{in}$). Car engines are roughly $15\%$ efficient.

8-9 Potential Energy Diagrams

• Potential Energy Diagram: A graph of $U(x)$ versus $x$. Total energy $E$ is a horizontal line.

• Turning Points: Points where $E = U(x)$, meaning kinetic energy $K = 0$. The object reverses direction.

• Bound Motion: If E < U_{max}, the object is trapped in a "potential well."

• Equilibrium Types:

  • Stable Equilibrium: Located at relative minima of $U(x)$. Displacement results in a restoring force back to equilibrium (e.g., Point $x_0$).

  • Unstable Equilibrium: Located at relative maxima of $U(x)$. Displacement results in a force pulling the object further away (e.g., Point $x_4$).

  • Neutral Equilibrium: Region where $U(x)$ is constant. Force remains zero when displaced (e.g., near $x_6$).

• ATP Storage: ATP bonds act as an energy store. Separation of molecules by catalysts lowers the potential energy barrier, releasing stored energy \Delta U = U(x_0) - U(\infty) > 0.

8-10 Gravitational Assist (Slingshot)

• Spacecraft can increase kinetic energy by passing near a moving planet.

• Reference Frames: In Jupiter's frame, spacecraft energy is symmetric (entry energy = exit energy). In the Solar System frame, the spacecraft can gain speed relative to the Sun.

• Calculation of Energy Change:

  • Final x-velocity $v'_{0x} = -v_{0x} + 2u$, where $u$ is planet velocity.

  • Change in kinetic energy: $\Delta K = 2mu(u - v_{0x})$ (Equation 8-23).

  • Increase occurs because initial $v_{0x}$ is negative (directed toward planet).

• Voyager 2: Launched in $1977$, utilized assists from Jupiter, Saturn, Uranus, and Neptune to enter interstellar space.