SCP1132 Introduction to Physics - Module 5: Work and Energy

Introduction

  • Course Details: SCP1132 Introduction to Physics

  • Module Focus: Work and Energy

  • Instructor: Associate Professor Steven Hinckley, School of Science

  • Text Reference: Cutnell & Johnson, Chapters 6 & 7.

Lecture Content Overview

  • Work

    • Physics Work vs Real Work

    • Units of Work

    • Work done by a force at an angle (to displacement)

  • Energy

    • Potential Energy and Gravitational Potential Energy

    • Kinetic Energy

    • Work-Kinetic Energy Theorem

    • Newton’s First Law (NFL)

  • Conservation of Energy

  • Power

  • Collisions and Momentum

    • Momentum

    • Conservation of Momentum

    • Types of Collisions: elastic, inelastic, perfectly inelastic

    • Impulse: Change of Momentum

Work

  • Definition of Work:

    • In physics, work is defined specifically as "The application of force over a displacement".

    • Mathematical Representation: W=FimessW = F imes s

  • Real Work:

    • Importance of distinguishing actual work from perceived effort:

    • Example: When displacement (d) is perpendicular to force, no work is done (net work = 0):
      Fimesextcos(90exto)=0F imes ext{cos}(90^ ext{o}) = 0

    • Constant horizontal velocity indicates no net acceleration and no net force in the horizontal direction.

Units of Work

  • A unit of work is defined as the work done to apply a force of 1 Newton (1 N = 1extkgimesextm/s21 ext{kg} imes ext{m/s}^2) through a displacement of 1 meter.

  • From the basic equation W=FimessW = F imes s, it follows:

    • Unit for work derived as W=extkgimesextm2/exts2W = ext{kg} imes ext{m}^2/ ext{s}^2 or NimesmN imes m.

  • This unit is named Joule (J), which is the same unit used for energy.

    • Historical Context: James Prescott Joule studied the relationship between heat and mechanical work, leading to the law of conservation of energy and foundational principles of thermodynamics.

Work Done by Forces at an Angle

  • When a force is applied at an angle to the direction of displacement, only the component of the force that is parallel contributes to the work done:

    • Work Equation: W=(Fimesextcos(heta))imess=Fsimesextcos(heta)W = (F imes ext{cos}( heta)) imes s = F_s imes ext{cos}( heta)

Example Calculation - Work Done by a Constant Force

  • Problem Statement: Find the work done if the force is 45.0 N, at an angle of 50.0 degrees, and displacement is 75.0 m.

    • Calculation Breakdown:

    • Work Done: W=Fimessimesextcos(heta)W = F imes s imes ext{cos}( heta)

    • Substitute values:
      W=(45.0extN)(75.0extm)extcos(50exto)W = (45.0 ext{ N})(75.0 ext{ m}) ext{cos}(50^ ext{o})

    • Result: Approximately Wext=2169extJext(orroundedto2170extJ,3significantfigures)W ext{ = } 2169 ext{ J} ext{ (or rounded to } 2170 ext{ J, 3 significant figures)}

Energy

  • Definition of Energy: The ability to do work.

  • When work is done on an object, it gains the ability to do work, hence acquiring energy.

  • Types of Energy:

    • Mechanical Energy is characterized into two categories:

    • Potential Energy: Energy stored in an object due to its position.

    • Kinetic Energy: Energy possessed by an object due to its motion.

Potential Energy

  • Types of Potential Energy:

    • Elastic Potential Energy: Energy stored in a stretched or compressed spring.

    • Chemical Potential Energy: Energy stored in chemical bonds.

    • Nuclear Potential Energy: Energy stored in atomic nuclei, especially in radioactive materials.

    • Electrical Potential Energy: Energy stored in batteries.

    • Gravitational Potential Energy (GPE): Most commonly observed in objects elevated above Earth’s surface.

Gravitational Potential Energy

  • Gravitational Potential Energy Formula:

    • PE=mghPE = mgh

    • Where:

      • m = mass (kg)

      • g = acceleration due to gravity (9.8extm/s29.8 ext{ m/s}^2)

      • h = height above ground (m)

  • Notation: Some texts may use the symbol U for GPE, often represented as extΔUext{Δ}U.

Activity: Potential Energy Calculation

  • Situation: A 1000 kg roller-coaster car moves between points with different heights.

    • (a) Calculate GPE at points 2 and 3, with point 1 as the reference point (y = 0).

    • At point 2:

      • PE2=mgh=(1000extkg)(9.8extm/s2)(10extm)=9.8imes104extJPE_2 = mgh = (1000 ext{ kg})(9.8 ext{ m/s}^2)(10 ext{ m}) = 9.8 imes 10^4 ext{ J}

    • At point 3:

      • PE3=mgh=(1000extkg)(9.8extm/s2)(−15extm)=−1.5imes104extJPE_3 = mgh = (1000 ext{ kg})(9.8 ext{ m/s}^2)(-15 ext{ m}) = -1.5 imes 10^4 ext{ J}

    • (b) Change in Potential Energy from 2 to 3:

    • extΔPE=PE<em>final−PE</em>initial=PE<em>3−PE</em>2=−1.5imes104extJ−9.8imes104extJ=−2.5imes105extJext{Δ}PE = PE<em>{final} - PE</em>{initial} = PE<em>3 - PE</em>2 = -1.5 imes 10^4 ext{ J} - 9.8 imes 10^4 ext{ J} = -2.5 imes 10^5 ext{ J}

    • (c) New Reference Point at 3, calculate GPE between points 1 and 2:

    • At point 1:

      • PE1=mgh=(1000extkg)(9.8extm/s2)(15extm)=+9.8imes104extJPE_1 = mgh = (1000 ext{ kg})(9.8 ext{ m/s}^2)(15 ext{ m}) = +9.8 imes 10^4 ext{ J}

    • At point 2:

      • PE2=mgh=(1000extkg)(9.8extm/s2)(25extm)=+2.5imes105extJPE_2 = mgh = (1000 ext{ kg})(9.8 ext{ m/s}^2)(25 ext{ m}) = +2.5 imes 10^5 ext{ J}

    • Change in Potential Energy:
      extΔPE=PE<em>final−PE</em>initial=0−2.5imes105extJ=−2.5imes105extJext{Δ}PE = PE<em>{final} - PE</em>{initial} = 0 - 2.5 imes 10^5 ext{ J} = -2.5 imes 10^5 ext{ J}

Kinetic Energy

  • Definition of Kinetic Energy (KE):

    • If an object is moving, it possesses energy due to that motion:

    • KE Formula: KE=rac12mv2KE = rac{1}{2}mv^2

      • Where m = mass (kg), v = velocity (m/s).

  • Unit of Kinetic Energy:

    • Derived from: KE=rac12mv2=extkgimesextm2/exts2=Nimesm=JKE = rac{1}{2}mv^2 = ext{kg} imes ext{m}^2/ ext{s}^2 = N imes m = J

Work-Energy Theorem

  • Relation Between Work and Kinetic Energy:

    • Gain in kinetic energy due to work done on an object when it speeds up.

    • Loss of kinetic energy due to work done when it slows down.

    • Work Equation:

    • W=extΔKEW = ext{Δ}KE

    • W=KE<em>2−KE</em>1=rac12mv<em>22−rac12mv</em>12W = KE<em>2 - KE</em>1 = rac{1}{2}mv<em>2^2 - rac{1}{2}mv</em>1^2

Activity: Work-KE Theorem Example

  • Problem: Calculate the net work required to accelerate a 1000 kg car from 20 m/s to 30 m/s.

    • Using the Work-KE Theorem:

    • W=extΔKEW = ext{Δ}KE

      • W=KE<em>2−KE</em>1=rac12(1000extkg)(30extm/s)2−rac12(1000extkg)(20extm/s)2W = KE<em>2 - KE</em>1 = rac{1}{2}(1000 ext{ kg})(30 ext{ m/s})^2 - rac{1}{2}(1000 ext{ kg})(20 ext{ m/s})^2

    • Result:
      W=4.5imes105extJ−2.0imes105extJ=2.5imes105extJW = 4.5 imes 10^5 ext{ J} - 2.0 imes 10^5 ext{ J} = 2.5 imes 10^5 ext{ J}

Conservation of Energy

  • Principle: Energy cannot be created or destroyed; it can only change forms.

  • Reference: Cutnell & Johnson, Section 6.5

Conservation of Energy in Closed Systems

  • Energy transformation occurs without the loss of total energy; hence:

    • Total energy remains constant in a closed system.

Activity: Energy Conservation Example

  • Scenario: Roller-coaster starts from rest at height 40 m.

    • Calculate the speed (v2) at the bottom of the hill (y2 = 0).

    • Applying conservation of energy:

    • Total energy at the top: E=mgh=0+mgyE = mgh = 0 + mgy

    • Bottom of the hill: E=rac12mv2+0E = rac{1}{2}mv^2 + 0

    • Therefore, equate: mgy=rac12mv2mgy = rac{1}{2}mv^2

      • Rearranging yields the final velocity equation:
        v=ext√(2gh)=ext√(2imes9.8extm/s2imes40extm)v = ext{√}(2gh) = ext{√}(2 imes 9.8 ext{ m/s}^2 imes 40 ext{ m})

      • Solving gives v = 28 m/s.

Activity: Daredevil Motorist

  • Scenario: A motorcyclist drives horizontally off a cliff at 38.0 m/s, neglecting air resistance.

    • Required: Find the speed upon striking the ground.

    • Change in potential energy relates to vertical displacement:

    • extΔU=mgimesextΔhext{Δ}U = mg imes ext{Δ}h

    • For conservation:

    • extΔK=rac12mv<em>f2−rac12mv</em>02ext{Δ}K = rac{1}{2}mv<em>f^2 - rac{1}{2}mv</em>0^2

    • Set expressions equal to find final velocity.

Power

  • Definition of Power: The rate at which work is done.

  • Units of Power:

    • P=J/sP = J/s or extJs−1ext{Js}^{−1}.

    • Renamed as Watt (W), defined as the rate of 1 Joule of work done in 1 second.

Historical Context of Power

  • Origin of the unit Watt: Named after Scottish inventor James Watt, who developed the steam engine crucial to the Industrial Revolution.

Power in Action

  • Example: The main engines of the space shuttle deliver 33,000 MW at take-off.

  • This high rate of power consumption equates to a fuel burn of 3400 kg/s, comparable to draining a swimming pool every 20 seconds.

Activity: Estimating Power Output

  • Scenario: A 60 kg jogger ascends a flight of stairs 4.5 m high in 4.0 s.

    • Energy expended:

    • extΔU=mgh=(60extkg)(9.8extm/s2)(4.5extm)=2646extJ=2600extJ(2s.f.)ext{Δ}U = mgh = (60 ext{ kg})(9.8 ext{ m/s}^2)(4.5 ext{ m}) = 2646 ext{ J} = 2600 ext{ J} (2 s.f.)

    • Power output:

    • P=racWextΔt=racmghextΔt=rac2646extJ4.0exts=660extWP = rac{W}{ ext{Δ}t} = rac{mgh}{ ext{Δ}t} = rac{2646 ext{ J}}{4.0 ext{ s}} = 660 ext{ W}

Momentum and Impulse

  • Definition of Momentum:

    • Known as the product of mass and velocity: p=mvp = mv.

  • Unit of Momentum: kg·m/s.

Impulse and Change of Momentum

  • When an unbalanced force is applied over a period of time, it results in a change in momentum:

    • Impulse (I): FimesextΔt=extChangeinMomentumF imes ext{Δ}t = ext{Change in Momentum}

Practical Application of Impulse

  • Safety Mechanisms: Airbags and seatbelts utilize the concept of impulse to reduce injury during collisions.

Activity: Calculating Average Force via Impulse

  • Scenario: A tennis ball with mass 0.06 kg leaves a racket at 55 m/s after contact for 4 ms.

  • Average force on the ball:

    • I=extΔp=p<em>f−p</em>iI = ext{Δ}p = p<em>f - p</em>i

    • Calculation:

    • I=mimes(v<em>f−v</em>i)=0.06extkgimes(55m/s−0)I = m imes (v<em>f - v</em>i) = 0.06 ext{ kg} imes (55 m/s - 0)

    • Resulting force: Calculate and check against weight of a 60 kg individual.

Collisions and Conservation of Momentum

  • Momentum remains constant in collisions.

  • In any system, the total momentum is unchanged:

    • p<em>initial=p</em>finalp<em>{initial} = p</em>{final}

Types of Collisions

  • Elastic Collision: Momentum and kinetic energy are conserved.

  • Inelastic Collision: Momentum is conserved, but kinetic energy is not.

  • Perfectly Inelastic Collision: Objects stick together after collision; momentum conserved but kinetic energy is not.

Activity: Momentum Calculation in Collisions

  • Scenario: A 10,000 kg railroad car A traveling at 24 m/s collides with a stationary identical car B, locking together.

  • Calculate speed post-collision:

    • Initial momentum = momentum of A (B is stationary).

    • Resulting speed = halved mass leads to halved speed after collision.

Summary of Key Concepts

  • Work:

    • Measured in Joules: W=FimessW = F imes s

  • Energy: Ability to do work; includes:

    • Gravitational Potential Energy: GPE=mghGPE = mgh

    • Kinetic Energy: KE=rac12mv2KE = rac{1}{2}mv^2

  • Conservation of Energy: Energy can neither be created nor destroyed, only transformed.

  • Power: Rate of work done: P=W/tP = W/t

  • Momentum: Defined as p=mvp = mv; conserved in isolated systems.

  • Impulse: Ft=extΔpFt = ext{Δp}

Upcoming Module 6 Topics

  • Concepts of Circular (Rotational) Motion:

    • Radial (Centripetal) Acceleration

    • Banked Curves

    • Torque

    • Centre of Mass

    • Static Equilibrium and Moments

  • Reference: Cutnell & Johnson, Chapters 5 and 9.

  • Tutorial 5: Discussion of tutorial questions for Module 5.