Conservation of Mechanical Energy

Defining the Law of Conservation

Mechanical energy ($E_{mech}$) is the sum of a system's kinetic energy and its potential energy. In an isolated system—where no external work is done by non-conservative forces like friction or air resistance—the total mechanical energy remains constant over time. This principle allows us to predict the state of a system (velocity, height, or compression) at any point in its motion without needing to calculate the specific forces at every interval.

Energy States and System Definitions

• Kinetic Energy ($K$): The energy of motion.

  • $K=\frac{1}{2}m{v}^2$
    

  • $m$ is mass (kg), $v$ is speed (m/s).

• Gravitational Potential Energy ($U_g$): Energy stored due to an object's position in a gravitational field.

  • $U_g=mgh$
    

  • $h$ is the vertical height relative to a defined "zero point" or reference level.

• Elastic (Spring) Potential Energy ($U_s$): Energy stored when a spring or elastic material is stretched or compressed.

  • $U_s=\frac{1}{2}k{x}^2$
    

  • $k$ is the spring constant (N/m), $x$ is the displacement from equilibrium (m).

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

• Closed vs. Open Systems:

  • Closed System: No energy enters or leaves. $E_{initial}=E_{final}$.

  • Open System: External work ($W$) is done on the system. $ΔE=W$.

• Common Pitfalls:

  • Sign of $x$: In $U_s$, $x$ is squared, so whether the spring is compressed or stretched, the potential energy is always positive.

  • Reference Level: Forgetting to define $h=0$ consistently throughout a problem. Usually, the lowest point in the motion should be $h=0$.

  • Internal vs. External: If the Earth is not included in your system, gravity is an "external force" doing work, rather than $U_g$ being internal energy. In AP Physics 1, always include the Earth in your system to use $U_g$.

Analyzing Real-World Scenarios

Scenario 1: The Roller Coaster (Gravitational to Kinetic) A 500 kg coaster car starts from rest at the top of a 40 m hill. How fast is it moving at the top of a loop that is 15 m high? (Assume no friction).

1. Identify initial and final states: $$E_i = U_g$$(since$v_i = 0$) $$E_f = K + U_g$$

2. Set $E_i=E_f$: $$mgh_i = \frac{1}{2}mv_f^2 + mgh_f$$

3. Cancel mass ($m$) and solve for $v_f$: $$gh_i = \frac{1}{2}v_f^2 + gh_f$$ $$(10)(40) = \frac{1}{2}v_f^2 + (10)(15)$$ $$400 = \frac{1}{2}v_f^2 + 150$$ $$250 = \frac{1}{2}v_f^2$$ $$500 = v_f^2$$ $$v_f \approx 22.36 \text{ m/s}$$

Scenario 2: The Horizontal Spring Launcher (Elastic to Kinetic) A 0.2 kg block is pushed against a spring ($k=500$ N/m), compressing it 0.1 m. The block is released on a frictionless surface. What is the block's speed when it leaves the spring?

1. Set $E_i=E_f$: $$U_s = K$$

2. Substitute equations: $$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$$

3. Cancel the $1/2$ and solve for $v$: $$(500)(0.1)^2 = (0.2)v^2$$ $$(500)(0.01) = 0.2v^2$$ $$5 = 0.2v^2$$ $$25 = v^2$$ $$v = 5 \text{ m/s}$$

Work and Non-Conservative Forces

The Work-Energy Theorem

Work ($W$) is the process of transferring energy into or out of a system via a force applied over a displacement. The Work-Energy Theorem states that the net work done on an object is equal to its change in kinetic energy: $W_{net}=ΔK$.

Components and Subtle Nuances

• Calculating Work: $W=Fd\cos (θ)$
  

  • $θ$ is the angle between the Force vector and the Displacement vector.

• Positive vs. Negative Work:

  • Positive Work: Force and displacement are in the same direction ($0^{∘}≤θ<90^{∘}$). Energy is added to the system.

  • Negative Work: Force and displacement are in opposite directions ($90^{∘}<θ≤180^{∘}$). Energy is removed (e.g., friction).

• Friction and Thermal Energy: When friction acts over a distance $d$, it does work $W_f=−f_{k}d$. This energy is not "lost" but converted into Internal Energy ($Q$ or $E_{th}$) (heat).

  • Conservation Equation with friction: $K_i+U_i+W_{nc}=K_f+U_f$ (where $W_{nc}$ is work by non-conservative forces).

• The "Area Rule": On a Force vs. Position ($F$ vs. $x$) graph, the work done is the area under the curve.

Friction and Work Application

Problem: A 2 kg box sliding at 10 m/s hits a rough patch of floor where the coefficient of kinetic friction ${μ}_k=0.5$. How far does it slide before coming to a stop?

1. Identify Work and Energy change: $$W_{friction} = \Delta K$$

2. Define $W_{friction}$ and $ΔK$: $$-f_k d = K_f - K_i$$ $$-(\mu_k mg)d = 0 - \frac{1}{2}mv^2$$

3. Cancel $m$ and solve for $d$: $$-(0.5)(10)d = -\frac{1}{2}(10)^2$$ $$-5d = -50$$ $$d = 10 \text{ meters}$$

Power: The Rate of Energy Transfer

Defining Power

Power ($P$) is the rate at which work is done or energy is transformed. In a physical sense, it measures how "fast" a task is completed.

Key Equations and Variations

• Average Power: $P=\frac{W}{Δt}=\frac{ΔE}{Δt}$
  

• Instantaneous Power: $P=Fv\cos (θ)$
  

  • This is derived from $P=\frac{Fd}{t}$ where $\frac{d}{t}=v$.

• Units: The Watt (W). $1\text{ W}=1\text{ J/s}=1\text{ kg}â‹{\text{m}}^2/{\text{s}}^3$.

• Nuance: If an object moves at a constant velocity while a force is applied (like a car driving against air resistance), the power delivered by the engine is exactly $P=F_{res}v$.

Power Calculation Example

An elevator motor lifts a 1000 kg cabin upward at a constant speed of 2 m/s. What is the power output of the motor?

1. Determine the force required:

   • Since speed is constant, $F_{lift}=F_g=mg$. $$F = (1000)(10) = 10,000 \text{ N}$$

2. Use the constant velocity power formula: $$P = Fv$$ $$P = (10,000 \text{ N})(2 \text{ m/s})$$ $$P = 20,000 \text{ Watts (or 20 kW)}$$

Energy Charts and Graphical Analysis

Visualizing Energy Transitions

AP Physics 1 frequently uses Energy Bar Charts (LOL Diagrams) and Energy vs. Position Graphs to test conceptual understanding of energy conservation.

Energy Bar Charts (LOL Diagrams)

• Left Side (L): Represents the initial energy ($K,U_g,U_s$).

• Middle Circle (O): Represents the defined system. Arrows pointing in/out represent Work ($W_{ext}$).

• Right Side (L): Represents the final energy.

• The Rule: Sum of bars on the left + Work in the circle = Sum of bars on the right.

Graphical Interpretation

• $F$ vs. $x$ Graph: Area = Work. graph[2x][0][10] In the graph above ($F=2x$), the work done from $x=0$ to $x=10$ is the area of the triangle: $\frac{1}{2}(10)(20)=100\text{ J}$.

• $U$ vs. $x$ Graph (Potential Energy Well):

  • The total energy ($E_{total}$) is a horizontal line.

  • The distance between the $U$ curve and the $E_{total}$ line is the Kinetic Energy ($K$).

  • Where the curve hits the $E_{total}$ line, $K=0$; these are "turning points."

Diagram Analysis Case Study

The Energy Chart Problem: A spring-loaded toy is compressed and then fires a ball straight up into the air.

• System: Ball, Spring, Earth.

• Initial State: Spring compressed, ball at rest.

  • Bars: Large $U_s$, zero $K$, zero $U_g$.

• Intermediate State (as it leaves the spring): Spring at equilibrium, ball moving.

  • Bars: Zero $U_s$, Large $K$, small $U_g$.

• Final State (at peak): Ball at highest point.

  • Bars: Zero $U_s$, zero $K$, Large $U_g$.

Key check: In a frictionless system, the total height of the bars in the first "L" must equal the total height of the bars in the last "L".