Young & Freedman University Physics - Chapter 07: Potential Energy & Conservation Study Guide

Overview of Energy Types

• Energy exists in many forms and is constantly transformed between types.

• Mechanical Energy ($ME$): Defined as the sum of Kinetic and Potential energy.

  • $ME = K + U$

• Non-Mechanical Energy ($NME$):

  • Thermal Energy: Energy associated with friction, heat, etc.

  • All Other Types: Includes electrical, nuclear, light, sound, etc. (covered in later studies).

• Kinetic Energy ($K$):

  • Energy due to motion.

  • Formula: $K = \frac{1}{2}mv^2$

• Potential Energy ($U$):

  • "Stored" energy due to position or configuration.

  • Elastic (Spring) Potential Energy ($U_{el}$): Energy stored due to compression or stretching ($x$) of a spring.

  • Gravitational Potential Energy ($U_g$): Energy stored due to height ($y$) relative to a reference level.

  • Formula: $U_g = mgy$

Conservation of Mechanical Energy

• Definition: The Mechanical Energy of a system is the sum of Kinetic ($K$) and Potential ($U$) energy.

• Conservatory Principle: When a system's $ME$ is transferred between potential and kinetic energy without loss, it is conserved.

  • Equation: $ME_i = ME_f$

  • Expanded: $K_i + U_i = K_f + U_f$

• Simplified Problem Solving Steps:

  1. Draw a diagram including initial ($i$) and final ($f$) states.

  2. Write the Conservation of Energy equation.

  3. Eliminate terms (e.g., if at rest, $K = 0$; if on the ground, $U = 0$) and expand remaining terms.

  4. Solve for the unknown.

• Uniformly Accelerated Motion (UAM) Equations (for comparison):

  1. $v = v_0 + at$

  2. $v^2 = v_0^2 + 2a\Delta x$

  3. $\Delta x = v_0t + \frac{1}{2}at^2$

  4. $\Delta x = \left(\frac{v + v_0}{2}\right)t$

Mechanical Energy Examples and Application

• Example 1: Dropping a Ball (ME Calculation)

  • Scenario: You drop a $2\,kg$ ball from the top of a $100\,m$ building.

  • Calculate $ME$ at the top: $K_i = 0$, $U_i = mgy = (2\,kg)(9.8\,m/s^2)(100\,m) = 1960\,J$. So, $ME = 1960\,J$ .

  • Calculate $ME$ right before hitting the ground: Since no energy is lost, $ME_f = 1960\,J$ (all $U$ converted to $K$).

• Example 2: Speed upon Impact

  • Scenario: A ball of unknown mass is dropped from $50\,m$.

  • Equation: $mgy_i = \frac{1}{2}mv_f^2$.

  • Mass ($m$) cancels out: $gy_i = \frac{1}{2}v_f^2$.

  • Result: $v_f = \sqrt{2gy_i} = \sqrt{2(9.8)(50)} \approx 31.3\,m/s$.

• Problem: Launching Upward

  • Scenario: A $4\,kg$ object is launched up from the ground with $v_0 = 40\,m/s$. Determine maximum height ($h_{max}$).

  • Equation: $K_i = U_f \rightarrow \frac{1}{2}mv_i^2 = mgh_{max}$.

  • Solve for $h_{max}$: $h_{max} = \frac{v_i^2}{2g} = \frac{40^2}{2(9.8)} \approx 81.6\,m$.

• Problem: Throwing Downward

  • Scenario: A $6\,kg$ object is thrown down from $20\,m$ and reaches the ground with $30\,m/s$. Calculate initial speed ($v_0$).

  • Equation: $K_i + U_i = K_f \rightarrow \frac{1}{2}mv_i^2 + mgy_i = \frac{1}{2}mv_f^2$.

• Multi-Point Problem Handling:

  • If a problem involves more than two points (e.g., launch, intermediate height, maximum height), select the two points that represent the "Given" (known values) and the "Target" (unknown value).

Conservation of Total Energy and Isolated Systems

• Total Energy ($E$): The sum of all types of energy (Mechanical + Non-Mechanical).

• Isolated System: A system where NO external forces do work; only internal forces do work.

  • System: A collection of objects chosen for analysis.

  • External Forces: Originate from outside the system. If they do work, the system is NOT isolated.

  • Internal Forces: Originate from objects within the system. If only internal forces do work, the system is isolated and total energy is conserved ($E_i = E_f$).

• Example: Spring Pushing a Box

  • System = Box Only: The force of the spring is EXTERNAL. The system is NOT isolated; work is done on the box, increasing its energy.

  • System = Box + Spring: The spring force is INTERNAL. The system IS isolated; energy is merely transferred from spring potential energy to box kinetic energy. Total energy remains constant.

Conservative vs. Non-Conservative Forces

• Conservative Forces: These forces are "reversible," meaning work done against them can be recovered (e.g., Gravity, Springs). $\Delta ME = 0$ if only these forces do work.

• Non-Conservative Forces ($W_{NC}$): These forces cause energy to be added to or removed from the mechanical system (e.g., Friction, Air Resistance, Applied Forces like a hand pushing).

• Situational Analysis:

  • Block falling (no air resistance): $ME$ is conserved; $U_g \rightarrow K$.

  • Moving block hits a spring and rebounds: $ME$ is conserved; $K \rightarrow U_{el} \rightarrow K$.

  • Person pushes a block from rest: $ME$ NOT conserved (Applied force work added); external energy converted to $K$.

  • Block slows due to friction: $ME$ NOT conserved ($K \rightarrow$ Thermal energy); mechanical energy is lost.

Energy Equation with Non-Conservative Forces

• When non-conservative forces act, we use the work-energy theorem expansion:

  • $K_i + U_i + W_{NC} = K_f + U_f$

• Work Formulas:

  • Total Work ($W$): $W = Fd \cos(\theta)$

  • Applied Force Work ($W_{F_A}$): $W_{F_A} = F_A d \cos(\theta)$

  • Kinetic Friction Work ($W_{f_k}$): $W_{f_k} = -f_k d$

• Problem: Hockey Puck

  • Scenario: $0.5\,kg$ puck at $4\,m/s$ is pushed with $200\,N$ for $0.3\,m$ on smooth ice.

  • Solve: $K_i + W_{F_A} = K_f \rightarrow \frac{1}{2}(0.5)(4)^2 + (200)(0.3) = \frac{1}{2}(0.5)v_f^2$.

• Problem: Sliding with Friction and Distance to Stop

  • Scenario: Block slides at $30\,m/s$ into a rough patch ($\mu_k = 0.6$). Calculate stopping distance.

  • Equation: $K_i + W_{f_k} = 0 \rightarrow \frac{1}{2}mv_i^2 - f_k d = 0$.

  • Substitute $f_k = \mu_k mg$: $\frac{1}{2}mv_i^2 - \mu_k mgd = 0$.

  • Result: $d = \frac{v_i^2}{2\mu_k g}$.

• Resistive Forces (Air Resistance):

  • Air resistance acts like friction. Work done by air resistance ($W_{air}$) is typically negative as it removes mechanical energy.

Specialized Energy Problems

• Curved Path Problems: If an object moves along complex, non-linear paths (e.g., a roller coaster or a smooth hill), conservation of energy is usually the only practical way to solve for speed/height.

  • Example: A block on a $20\,m$ high smooth hill requires a minimum speed at the bottom such that $K_i = U_f \rightarrow v_i = \sqrt{2gh}$.

• Connected Systems: When solving for multiple objects connected by strings/pulleys:

  • Consider the energies ($K$ and $U$) of EACH object ($K_{tot} = K_1 + K_2$, etc.).

  • Connected objects move with the same magnitude of velocity ($|v|$) and acceleration ($a$).

  • Example: A $5\,kg$ block hangs higher than a $4\,kg$ block; as it drops, it loses $U_g$, while both blocks gain $K$.

• Projectile Motion: Some projectile problems can be solved faster with energy if internal direction/components are not required.

  • Example: Ball thrown from $30\,m$ at $20\,m/s$ at an unknown angle. Final speed is independent of the angle because $K_i + U_i = K_f$.

Elastic Potential Energy

• Spring Potential Energy Formula: $U_{el} = \frac{1}{2}kx^2$.

• Work by Spring ($W_s$): $W_s = -\Delta U_{el} = -\frac{1}{2}kx^2$.

• Problem Strategy:

  • Stationary/Static problems: Use Force equations ($\sum F = 0$).

  • Moving objects: Use Energy Conservation because the spring force is not constant ($F_s = -kx$).

• Conservation Equation containing Springs: $K_i + U_{gi} + U_{eli} + W_{NC} = K_f + U_{gf} + U_{elf}$.

• Example: Block Hitting a Spring

  • Frictionless scenario ($W_{NC} = 0$): All $K$ converts to $U_{el}$. $\frac{1}{2}mv^2 = \frac{1}{2}kx_{max}^2$.

  • With Friction scenario ($W_{NC} = -f_k x$): $\frac{1}{2}mv^2 - f_k x = \frac{1}{2}kx^2$. This requires solving a quadratic equation for $x$.

Potential Energy Graphs ($U(x)$)

• Graphing: $\text{y-axis} = U(x)$, $\text{x-axis} = \text{position (x)}$.

• Mechanical Energy Line: Total $ME$ is a horizontal line of constant value (if $W_{NC} = 0$).

• Kinetic Energy on Graph: The vertical distance between the horizontal $ME$ line and the $U(x)$ curve.

  • $K = ME - U(x)$.

• Turning Points: Locations where $ME = U(x)$. At these points, $K = 0$, and the object reverses direction. The object is "trapped" between turning points if it lacks sufficient energy to cross over a peak.

• Forces and Slope:

  • Force is the negative slope of the potential energy graph: $F(x) = -\frac{dU}{dx}$.

  • Negative slope ($U(x)$ goes down): $F$ is positive (points toward increasing $x$).

  • Positive slope ($U(x)$ goes up): $F$ is negative (points toward decreasing $x$).

  • Zero slope ($U(x)$ is flat): $F = 0$, representing Equilibrium.

• Equilibrium Types:

  • Stable Equilibrium: Located at a local minimum of $U(x)$. The graph curves up ($\cup$). If nudged, restoring forces return the object to the minimum.

  • Unstable Equilibrium: Located at a local maximum of $U(x)$. The graph curves down ($\cap$). If nudged, the object moves further away.