AP Physics 1Work, Energy, and Power

Work & Energy Intro

## System Analysis & Energy Classifications *(Essential for Multiple Choice & Ranking Tasks)* ### Definitions & System Boundaries * **System:** A defined collection of objects. The choice of system determines how energy transfer is classified (Work vs. Potential Energy). * **Closed System:** No mass or energy enters or leaves. Total Energy is conserved ($\Delta E_{total} = 0$). * **Open System:** External forces perform Work on the system, changing its total energy ($W_{ext} = \Delta E$). * **Kinetic Energy ($K$):** Energy of motion. $K = \frac{1}{2}mv^2$. Scalar quantity (always positive). * **Gravitational Potential Energy ($U_g$):** Energy stored due to position in a gravitational field. $U_g = mgh$ (near Earth). Requires the Earth to be part of the system. * **Elastic Potential Energy ($U_s$):** Energy stored in a deformed elastic object (spring). $U_s = \frac{1}{2}kx^2$. ### Key Relationships & Proportionality (Ranking Task Logic) * **Velocity vs. Kinetic Energy:** Since $K \propto v^2$, doubling velocity quadruples kinetic energy. * *Ranking Tip:* When ranking objects by energy, check the square of the velocity first; it has a larger impact than mass. * **Height vs. Gravitational Energy:** $U_g \propto h$. Linear relationship. * **Mechanical Energy ($ME$):** The sum of kinetic and potential energies ($ME = K + U$). Conserved only if no non-conservative forces (friction, air resistance) do work. ### Conceptual Application: Dropped Object **Scenario:** A block of mass $m$ is dropped from height $h$. * **System = Block only:** Gravity is an *external* force. Gravity does positive work on the block, increasing its Kinetic Energy. * **System = Block + Earth:** Gravity is an *internal* force. Potential Energy ($U_g$) transforms into Kinetic Energy ($K$). Total Mechanical Energy remains constant. --- ## The Work-Energy Theorem *(Crucial for FRQs and Calculation Problems)* ### Force-Displacement Definition * **Work ($W$):** The transfer of energy via a force acting over a distance. $$W = F d \cos(\theta)$$ * $F$: Magnitude of Force. * $d$: Magnitude of Displacement. * $\theta$: Angle between the Force vector and Displacement vector (tail-to-tail). * **Work-Energy Theorem:** The net work done on an object equals its change in kinetic energy. $$W_{net} = \Delta K = K_f - K_i$$ ### Nuances & Graphical Analysis * **Sign Convention:** * **Positive Work ($0^\circ \leq \theta < 90^\circ$):** Force adds energy; object speeds up (if $F_{net}$). * **Negative Work ($90^\circ < \theta \leq 180^\circ$):** Force removes energy; object slows down (e.g., Friction). * **Zero Work ($\theta = 90^\circ$):** Force is perpendicular to motion (e.g., Normal Force on a flat surface, Tension in a pendulum). * **Force vs. Position Graphs:** * The **area under the curve** of an $F$ vs. $x$ graph represents Work done. * graph[x][0][10] (Imagine a linear force here; the area under the line is the work). ### Worked Example: Box on a Rough Surface **Problem:** A 10 kg box is pulled 5 meters across a rough floor by a tension force of 50 N at an angle of $37^\circ$ above horizontal. The coefficient of kinetic friction $\mu_k$ is 0.2. Calculate the final velocity if it starts from rest. 1. **Calculate Vertical Forces (to find Normal Force $F_N$):** * $\sum F_y = 0 \Rightarrow F_N + T\sin(37^\circ) - mg = 0$ * $F_N = mg - T\sin(37^\circ)$ * $F_N = (10)(9.8) - 50(0.6)$ * $F_N = 98 - 30 = 68 \text{ N}$ 2. **Calculate Friction Force ($f_k$):** * $f_k = \mu_k F_N$ * $f_k = 0.2(68) = 13.6 \text{ N}$ 3. **Calculate Work done by individual forces:** * $W_{tension} = T d \cos(37^\circ) = 50(5)(0.8) = 200 \text{ J}$ * $W_{friction} = f_k d \cos(180^\circ) = 13.6(5)(-1) = -68 \text{ J}$ * $W_{gravity} = 0 \text{ J}$ and $W_{normal} = 0 \text{ J}$ (Perpendicular to motion). 4. **Apply Work-Energy Theorem:** * $W_{net} = W_{tension} + W_{friction} = 200 - 68 = 132 \text{ J}$ * $W_{net} = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$ * $132 = \frac{1}{2}(10)v_f^2 - 0$ * $26.4 = v_f^2$ * $v_f \approx 5.14 \text{ m/s}$ --- ## Hooke's Law & Spring Potential Energy *(Focus of Derivations and Relationships)* ### Hooke's Law (Restoring Force) * **Definition:** The force exerted by a spring is proportional to its displacement from equilibrium and acts in the opposite direction. $$F_s = -k x$$ * $k$: Spring Constant (N/m). Measure of stiffness (higher $k$ = stiffer spring). * $x$: Displacement from equilibrium position ($x=0$). * The negative sign indicates it is a **restoring force**. * **Linear Relationship:** * On a Force ($y$-axis) vs. Displacement ($x$-axis) graph, the relationship is linear. * **Slope = Spring Constant ($k$).** * graph[2x][-5][5] (Example of linear Hooke's Law relationship). ### Elastic Potential Energy ($U_s$) * **Derivation from Graph:** * Work is the area under the Force vs. Displacement graph. * For a spring, the graph is a triangle with base $x$ and height $kx$. * $\text{Area} = \frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}(x)(kx)$. * **Formula:** $$U_s = \frac{1}{2} k x^2$$ * **Parabolic Relationship:** * Energy is proportional to the *square* of displacement. * Doubling the stretch distance ($x$) quadruples the stored energy ($U_s$). * graph[x^2][-5][5] (Example of parabolic Energy vs. Displacement). ### Worked Example: The Pinball Launcher (FRQ Style) **Problem:** A spring with constant $k = 400 \text{ N/m}$ is compressed by $0.1 \text{ m}$. A ball of mass $0.05 \text{ kg}$ is placed against it. The spring is released, launching the ball across a frictionless horizontal surface. 1. **State Conservation Principle:** * $E_{initial} = E_{final}$ (No non-conservative work). * $U_{s,i} + K_i = U_{s,f} + K_f$ 2. **Substitute Variables:** * Initial: Spring compressed ($x_i$), Ball at rest ($v_i=0$). * Final: Spring relaxed ($x_f=0$), Ball moving ($v_f$). * $\frac{1}{2}k x^2 + 0 = 0 + \frac{1}{2}m v^2$ 3. **Solve for Velocity:** * $k x^2 = m v^2$ * $v = \sqrt{\frac{k x^2}{m}}$ * $v = x \sqrt{\frac{k}{m}}$ * $v = 0.1 \sqrt{\frac{400}{0.05}}$ * $v = 0.1 \sqrt{8000}$ * $v = 0.1 (89.44)$ * $v \approx 8.94 \text{ m/s}$ ### Common Derivation/Ranking Trap * **Scenario:** Two different springs (Stiff $k_1$ and Loose $k_2$) are compressed by the *same force* $F$. Which stores more energy? * **Analysis:** * $F = kx \Rightarrow x = F/k$ * $U_s = \frac{1}{2}kx^2 = \frac{1}{2}k(\frac{F}{k})^2 = \frac{F^2}{2k}$ * **Result:** Energy is *inversely* proportional to $k$ when Force is constant. The looser spring (smaller $k$) stores more energy because it displaces further for the same force.

System Analysis &amp; Energy Classifications(Essential for Multiple Choice &amp; Ranking Tasks)Definitions &amp; System BoundariesSystem: A defined collection of objects. The choice of system determines how energy transfer is classified (Work vs. Potential Energy).Closed System: No mass or energy enters or leaves. Total Energy is conserved (ΔEtotal​=0).Open System: External forces perform Work on the system, changing its total energy (Wext​=ΔE).Kinetic Energy (K): Energy of motion. K=21​mv2. Scalar quantity (always positive).Gravitational Potential Energy (Ug​): Energy stored due to position in a gravitational field. Ug​=mgh (near Earth). Requires the Earth to be part of the system.Elastic Potential Energy (Us​): Energy stored in a deformed elastic object (spring). Us​=21​kx2.Key Relationships &amp; Proportionality (Ranking Task Logic)Velocity vs. Kinetic Energy: Since K∝v2, doubling velocity quadruples kinetic energy.Ranking Tip: When ranking objects by energy, check the square of the velocity first; it has a larger impact than mass.Height vs. Gravitational Energy: Ug​∝h. Linear relationship.Mechanical Energy (ME): The sum of kinetic and potential energies (ME=K+U). Conserved only if no non-conservative forces (friction, air resistance) do work.Conceptual Application: Dropped ObjectScenario: A block of mass m is dropped from height h.System = Block only: Gravity is an external force. Gravity does positive work on the block, increasing its Kinetic Energy.System = Block + Earth: Gravity is an internal force. Potential Energy (Ug​) transforms into Kinetic Energy (K). Total Mechanical Energy remains constant.The Work-Energy Theorem(Crucial for FRQs and Calculation Problems)Force-Displacement DefinitionWork (W): The transfer of energy via a force acting over a distance.W=Fdcos(θ)F: Magnitude of Force.d: Magnitude of Displacement.θ: Angle between the Force vector and Displacement vector (tail-to-tail).Work-Energy Theorem: The net work done on an object equals its change in kinetic energy.Wnet​=ΔK=Kf​−Ki​Nuances &amp; Graphical AnalysisSign Convention:Positive Work (0∘≤θ&lt;90∘): Force adds energy; object speeds up (if Fnet​).Negative Work (90∘&lt;θ≤180∘): Force removes energy; object slows down (e.g., Friction).Zero Work (θ=90∘): Force is perpendicular to motion (e.g., Normal Force on a flat surface, Tension in a pendulum).Force vs. Position Graphs:The area under the curve of an F vs. x graph represents Work done.graph[x][0][10] (Imagine a linear force here; the area under the line is the work).Worked Example: Box on a Rough SurfaceProblem: A 10 kg box is pulled 5 meters across a rough floor by a tension force of 50 N at an angle of 37∘ above horizontal. The coefficient of kinetic friction μk​ is 0.2. Calculate the final velocity if it starts from rest.Calculate Vertical Forces (to find Normal Force FN​):∑Fy​=0⇒FN​+Tsin(37∘)−mg=0FN​=mg−Tsin(37∘)FN​=(10)(9.8)−50(0.6)FN​=98−30=68 NCalculate Friction Force (fk​):fk​=μk​FN​fk​=0.2(68)=13.6 NCalculate Work done by individual forces:Wtension​=Tdcos(37∘)=50(5)(0.8)=200 JWfriction​=fk​dcos(180∘)=13.6(5)(−1)=−68 JWgravity​=0 J and Wnormal​=0 J (Perpendicular to motion).Apply Work-Energy Theorem:Wnet​=Wtension​+Wfriction​=200−68=132 JWnet​=ΔK=21​mvf2​−21​mvi2​132=21​(10)vf2​−026.4=vf2​vf​≈5.14 m/sHooke's Law &amp; Spring Potential Energy(Focus of Derivations and Relationships)Hooke's Law (Restoring Force)Definition: The force exerted by a spring is proportional to its displacement from equilibrium and acts in the opposite direction.Fs​=−kxk: Spring Constant (N/m). Measure of stiffness (higher k = stiffer spring).x: Displacement from equilibrium position (x=0).The negative sign indicates it is a restoring force.Linear Relationship:On a Force (y-axis) vs. Displacement (x-axis) graph, the relationship is linear.Slope = Spring Constant (k).graph[2x][-5][5] (Example of linear Hooke's Law relationship).Elastic Potential Energy (Us​)Derivation from Graph:Work is the area under the Force vs. Displacement graph.For a spring, the graph is a triangle with base x and height kx.Area=21​(base)(height)=21​(x)(kx).Formula: Us​=21​kx2Parabolic Relationship:Energy is proportional to the square of displacement.Doubling the stretch distance (x) quadruples the stored energy (Us​).graph[x^2][-5][5] (Example of parabolic Energy vs. Displacement).Worked Example: The Pinball Launcher (FRQ Style)Problem: A spring with constant k=400 N/m is compressed by 0.1 m. A ball of mass 0.05 kg is placed against it. The spring is released, launching the ball across a frictionless horizontal surface.State Conservation Principle:Einitial​=Efinal​ (No non-conservative work).Us,i​+Ki​=Us,f​+Kf​Substitute Variables:Initial: Spring compressed (xi​), Ball at rest (vi​=0).Final: Spring relaxed (xf​=0), Ball moving (vf​).21​kx2+0=0+21​mv2Solve for Velocity:kx2=mv2v=mkx2​​v=xmk​​v=0.10.05400​​v=0.18000​v=0.1(89.44)v≈8.94 m/sCommon Derivation/Ranking TrapScenario: Two different springs (Stiff k1​ and Loose k2​) are compressed by the same force F. Which stores more energy?Analysis:F=kx⇒x=F/kUs​=21​kx2=21​k(kF​)2=2kF2​Result: Energy is inversely proportional to k when Force is constant. The looser spring (smaller k) stores more energy because it displaces further for the same force.