(6/10/26) Force, Work, Energy, and Power

Course Logistics and Exam Administration

• Exam 2 Overview

  • The second exam will cover the topics of Force, Work, Energy, and Power.
  • Equation Sheets: Students will be provided with an equation sheet similar to the checklist provided for the first exam. The instructor states that there is no need to memorize basic equations by heart, as they will be included.
  • Necessary Basic Knowledge: While equations are provided, students are expected to recognize basic formulas such as gravitational force ($F_g = m \times g$) and work ($W = F \times d \times \text{cos}(\theta)$).
  • Exam Structure: The format remains consistent with the first exam, consisting of four problems in Part 1 and four problems in Part 2.
  • Duration: The exam time is approximately one hour and fifteen minutes.

• Final Exam Structure

  • The cumulative weight of the final exam is approximately 20% material from before Exam 2 and 80% material covered after Exam 2.
  • Topics like momentum, torque, and rotational motion will feature heavily in the final.

• Curriculum Adjustments regarding Friction

  • The instructor will dedicate a full class to teaching friction and will upload numerous problems with solutions.
  • By student consensus, complex friction problems (such as comparing multiple traveling cars) will be minimized on the second exam in favor of basic conceptual multiple-choice questions (e.g., comparing momentum levels).
  • Friction is highlighted as a critical topic that will be used extensively in Physics II.

Power: Definition and Mathematical Expressions

• Definition of Power

  • Power is defined as the rate at which energy is used or the rate at which work is performed.
  • It represents how fast work gets done; an instrument with higher power can complete more work in less time than one with lower power.

• Fundamental Equations

  • The primary definition based on work and time:     $P = \frac{W}{t}$
  • Since $W = F \times d$, power can also be expressed in terms of force and velocity:     $P = F \times v$
  • Where $v$ is velocity ($\frac{d}{t}$).

• Standard Units of Power

  • Watt (W): The standard SI unit, defined as one Joule per second ($1\,W = 1\,J/s$).
  • Horsepower (HP): Commonly used for engines. The conversion factor is:     $1\,HP = 745\,W$

• Dimensional Expression of Units     The instructor identifies three ways to express the units of power:

  1. Joules per second ($J/s$).
  2. Newton-meters per second ($N \cdot m/s$).
  3. Kilogram-meters squared per second cubed ($kg \cdot m^2/s^3$).
     • This is derived from $P = m \times a \times v$, resulting in $kg \times (m/s^2) \times (m/s) = kg \cdot m^2/s^3$.

Scientific Notation and Prefixes

• Large Scale (Positive Powers of 10)

  • Kilo (k): $10^3$ (e.g., $1000\text{ meters} = 1\text{ kilometer}$).
  • Mega (M): $10^6$.
  • Giga (G): $10^9$ (often used for space vehicles or rockets).

• Small Scale (Negative Powers of 10)

  • Milli (m): $10^{-3}$.
  • Micro (\mu): $10^{-6}$.
  • Nano (n): $10^{-9}$.
  • Fermi/Pico: The transcript mentions Fermi at $10^{-12}$ and $10^{-15}$.

The Work-Energy Theorem

• Core Principle

  • Work and Energy are equivalent and scalar quantities.
  • Individual works performed on a system can be added arithmetically: $W_{total} = W_1 + W_2 + W_3 + \dots$

• Theorem Definition

  • The theorem states that the change in kinetic energy ($\Delta KE$) of a system is equal to the sum of the work done by all individual forces acting upon the system.
  • Equation: $\Delta KE = KE_f - KE_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

Case Study: Hydroelectric Plants

• Mechanism of Action

  • Electricity is produced from the potential energy of water stored at a height (behind a dam).
  • Water is projected through a concentrated nozzle to rotate a turbine, converting potential energy into kinetic (rotational) energy, and finally electrical energy.

• Numerical Example

  • Flow rate: $2000\,m^3/s$.
  • Mass of water: $1\,m^3$ of water is approximately $1000\,kg$.
  • Weight: $1000\,kg$ corresponds to roughly $9800\,N$.
  • Height: Falling $20\,m$ can yield around $400\,MW$ (megawatts) of power.

Step-by-Step Problem Solving

• Scenario 1: Personal Power Output (Running Upstairs)

  • Parameters: Mass ($m$) = $70\,kg$, Height ($h$) = $3\,m$, Time ($t$) = $3\,s$, Gravity ($g$) = $10\,m/s^2$.
  • Force calculation: $F = m \times g = 70 \times 10 = 700\,N$.
  • Velocity calculation: $v = \frac{h}{t} = \frac{3\,m}{3\,s} = 1\,m/s$.
  • Power calculation: $P = F \times v = 700\,N \times 1\,m/s = 700\,W$.

• Scenario 2: Pushing a Cart

  • Parameters: Mass = $5\,kg$, Force = $30\,N$, Distance = $10\,m$, Time = $5\,s$.
  • Velocity Calculation: $v = \frac{10\,m}{5\,s} = 2\,m/s$.
  • Power Calculation: $P = F \times v = 30\,N \times 2\,m/s = 60\,W$.

• Scenario 3: Particle Velocity and Kinetic Energy Change

  • Parameters: Mass = $2\,kg$, Initial Velocity ($v_i$) = $5\,m/s$, Final Velocity ($v_f$) = $10\,m/s$.
  • Calculation:
    1. Initial Kinetic Energy ($KE_i$): $\frac{1}{2} \times 2 \times 5^2 = 25\,J$.
    2. Final Kinetic Energy ($KE_f$): $\frac{1}{2} \times 2 \times 10^2 = 100\,J$.
    3. Change in Kinetic Energy ($\Delta KE$): $100\,J - 25\,J = 75\,J$.
  • This $75\,J$ is equivalent to the total work done on the ball.

• Scenario 4: The Waterfall Power Generator

  • Parameters: Height ($h$) = $948\,m$, Mass ($m$) = $6 \times 10^4\,kg$ per minute, Efficiency = 80%.
  • Potential Energy Calculation: $PE = m \times g \times h$.
  • Energy at Bottom: By the conservation of energy, the gravitational potential energy at the top converts entirely to kinetic energy at the bottom ($m \times g \times h = \frac{1}{2}mv^2$).
  • Velocity at Bottom: $v = \sqrt{2gh}$.
  • Output Power: Calculate 80% of the total potential energy divided by the time (60 seconds) to find the electrical power output.

Dimensional Analysis

• Utility: Dimensional analysis serves as a tool to verify the validity of derived equations or to eliminate incorrect multiple-choice options before solving a problem.
• Methodology: Align the units of each variable (Mass ($M$), Length ($L$), Time ($T$)) to see if the final units match the expected physical quantity (e.g., the units of Power).

Questions & Discussion

• Student: On the first exam, we had that checklist/equation sheet. What will we get on the second exam?
• Instructor: The second exam will also provide whatever equations you need. There wasn't a single equation you needed to learn by heart for the first one, and it will be the same for the next.
• Student: For vector multiplication [cross product/dot product], is that also going to be on it?
• Instructor: I might give you vectors and tell you to multiply them, but it’s more critical for torque, which we will do after momentum. When we learn torque, the cross product will return.
• Instructor: Do you want me to have friction in the exam?
• Student: I'd say if we're not going to be using it too much until Physics II, we can leave it out.
• Instructor: You are going to use friction a lot. I will include it but at a basic multiple-choice level, not complex comparison problems.
• Student: How do you rate the difficulty of the first exam out of 10?
• Student 1: Maybe a 5 out of 10.
• Student 2: I’d say a 7, it was fair, I just wasn't prepared.
• Instructor: 7 is good; I’ll keep that in mind. I was trying to make it easier, but for the final, the structure will be different.
• Student: Can we move on to what you want to teach today regarding Power?
• Instructor: Yes, let's come back to Power.
• Student: Will the final be cumulative?
• Instructor: The weightage will be around 20% before [Exam 2] and 80% after. We will decide for sure and get back to you.