Work, Energy, and Simple Machines Study Guide

Introduction to Work, Energy, and Simple Machines

• Foundational Context:

  • Kinematic equations and Newton's laws of motion describe how forces alter the motion of objects.
  • When forces vary with time or act in complex geometric arrangements, applying Newton's laws directly becomes computationally difficult.
  • The concepts of work, energy, and power provide a simplified and universal framework for analyzing motion and physical interactions.

• Role of Energy in Daily Life:

  • Energy is defined as the capacity to do work.
  • Everyday tasks rely on distinct energy sources:
  • Food: Provides biological chemical energy required for humans to walk and execute physical tasks.
  • Electricity: Provides electrical energy to drive electric motors and rotate ceiling fans.
  • Fuel: Provides chemical thermal energy to power internal combustion engines and propel automobiles.

• Simple Machines:

  • Simple machines serve as fundamental mechanical building blocks that allow tasks to be performed with reduced effort or enhanced convenience by altering force magnitude or direction.

• Think It Over (Playground Slide Dynamics):

  • Slide velocity magnitude: Determined by the conversion of gravitational potential energy at the top into kinetic energy at the bottom.
  • Mass independence: Assuming negligible friction, children of different masses reach the bottom of the same slide with the exact same velocity magnitude.
  • Slide geometry: The slide with the greatest vertical height results in the largest final velocity magnitude at the bottom, regardless of path curvature.

Work Done by a Constant Force

• Qualitative Conceptualization via Bag-Lifting:
  • Single Bag ($5\,\text{kg}$ to $1\,\text{m}$):
  • Downward gravitational force: $F_g = mg$, where $m = 5\,\text{kg}$ and $g$ is acceleration due to gravity.
  • Minimum upward lifting force required: $F = mg$.
  • Vertical displacement: $s = 1\,\text{m}$ in the direction of applied force.
  • Mechanical work performed is designated as $W$.
  • Sequential Lifting (3 bags, $1\,\text{m}$ each):
  • Lifting three $5\,\text{kg}$ bags sequentially to a height of $1\,\text{m}$ requires doing work equal to $3W$.
  • A fuel-powered machine performing this task requires 3 times more fuel.
  • Simultaneous Lifting (3 bags together to $1\,\text{m}$):
  • Total mass: $3m = 15\,\text{kg}$.
  • Required upward force: $3F = 3mg$
  • Total work done: $3W$. Applying a larger force over the same distance produces a proportionally larger amount of work.
  • Increased Height (1 bag to $3\,\text{m}$):
  • Constant applied force: $F = mg$
  • Vertical displacement: $s = 3\,\text{m}$.
  • Total work done: $3W$. Applying the same force over a larger displacement produces a proportionally larger amount of work.

• Verbatim Definition of Work Done:   $\text{work done on an object by a constant force} = \text{force applied} \times \text{displacement in the direction of the force}$

• General Mathematical Formula:

  • For a constant force $F$ acting on an object undergoing displacement $s$ parallel to the direction of force:   $W = F \times s$

• Specification Rule:

  • When describing work, it is mandatory to specify both the force (or agency) doing the work and the object upon which the work is performed.

• SI Unit and Dimensional Derivation:

  • The SI unit of work is the joule ($\text{J}$).
  • One joule is defined as:   $1\,\text{J} = 1\,\text{N} \times 1\,\text{m}$
  • One joule of work is done on an object when a constant force of 1 newton displaces it by 1 metre in the direction of the force.
  • Dimensional analysis in base SI units:   $1\,\text{N} = 1\,\text{kg\,m\,s}^{-2}$$1\,\text{J} = 1\,\text{kg\,m\,s}^{-2} \times 1\,\text{m} = 1\,\text{kg\,m}^2\,\text{s}^{-2}$

• Force-Displacement ($F-s$) Graph Analysis:

  • Plotting force $F$ on the Y-axis against displacement $s$ on the X-axis yields a curve whose area represents mechanical work.
  • For a constant force of $10\,\text{N}$ displacing an object by $1.0\,\text{m}$, the work done equals the shaded rectangular area:   $W = 10\,\text{N} \times 1.0\,\text{m} = 10\,\text{J}$
  • For variable forces, work done is calculated by determining the total area under the $F-s$ curve between initial position $s_i$ and final position $s_f$

• Conditions for Zero Work Done ($W = 0$):
  1. Zero Force ($F = 0$): No work is done if no force acts on the object.
  2. Zero Displacement ($s = 0$): Pushing against a rigid, stationary wall exerts force, but because displacement is zero ($s = 0$), no mechanical work is done on the wall.
  • Physiological Fatigue: Internal energy is consumed by muscle fibers expanding and contracting repeatedly to maintain force, causing tiredness despite zero mechanical work on the target object.
  1. Perpendicular Force and Displacement ($F \perp s$): When applied force acts perpendicular to displacement, zero work is performed by that force because displacement in the force's direction is zero.
  • Example: A girl carrying a heavy box horizontally while walking at constant speed applies an upward vertical force to balance weight. Since force is vertical and displacement is horizontal, work done by her lifting force on the box is zero.

• Positive vs. Negative Work Done:
  • Work is a scalar quantity (has magnitude but no directional vector component), but can be positive or negative based on alignment.
  • Positive Work: Applied force and displacement are in the same direction. Example: Pushing a wheelchair forward increases its speed.
  • Negative Work: Applied force and displacement are in opposite directions. Example: A goalkeeper applying force to decelerate and stop a moving football performs negative work on the ball.

• Interaction Pair Perspective (Newton's Third Law):

  • When a goalkeeper stops a ball, equal and opposite forces act on both objects.
  • The ball exerts force on the goalkeeper in the direction of motion, doing positive work on the goalkeeper.
  • The goalkeeper exerts opposing force on the ball, doing negative work on the ball.

• Worked Examples:

  • Example 7.1 (Dumbbell Exercise):
  • Lifting dumbbell upward: Upward force aligns with upward displacement $\rightarrow$ Positive Work.
  • Lowering dumbbell slowly: Upward holding force opposes downward displacement $\rightarrow$ Negative Work.
  • Example 7.2 (Goalkeeper Stopping Ball):
  • Force $F = 200\,\text{N}$, hand displacement $s = -15\,\text{cm} = -0.15\,\text{m}$ (opposite to force).     $W = F \times s = 200\,\text{N} \times (-0.15\,\text{m}) = -30\,\text{J}$

• Pause and Ponder Answers:

  1. Weightlifter holding barbell steady: Zero displacement ($s=0$) means work done on the barbell is zero.
  2. Friction on sliding coins: Friction acts opposite to sliding displacement, so work done by friction is negative.

The Work-Energy Theorem

• Energy Acquisition via Work:
  • Doing positive work on an object imparts energy to it, giving it the capacity to do work on other objects.
  • Example 1: A fielder does work to throw a cricket ball. The moving ball possesses energy and knocks down wickets upon impact, transferring energy to them.
  • Example 2: Raising a flowerpot to a balcony height requires work. If dropped, the falling pot can damage objects below.

• Work-Energy Theorem Statement:

  • Mechanical work done on an object equals the net change in its energy:   $\text{work done on an object} = \text{change in its energy}$$W = \Delta E$
  • Applies to single particles, rigid bodies, complex systems, and variable force fields.
  • The SI unit of energy is identical to work: joule ($\text{J}$).

• Non-Mechanical Energy Transfer Mechanisms:

  • Mechanical work is one mode of energy transfer. Other fundamental modes include:
  1. Heat Conduction: Thermal energy flows across temperature gradients upon direct contact.
  2. Thermal Radiation: Electromagnetic waves transfer energy across vacuum (e.g., Solar radiation reaching Earth).
  3. Electric Circuits: Electrical potential differences drive charge flow.
  4. Sound Waves: Mechanical pressure waves transfer acoustic energy through media.
  5. Nuclear Reactions: Mass-energy conversion powering stars.

• Meet a Scientist: James Prescott Joule:

  • Investigated the quantitative equivalated relationship between mechanical energy and thermal energy, establishing the First Law of Thermodynamics and unifying the concept of energy.

• Worked Example 7.3 (Carrom Board Collisions):
  • Striker hits white coin: Striker exerts force along white coin's displacement $\rightarrow$ positive work on white coin (increases white coin's kinetic energy). White coin exerts equal opposite force on striker $\rightarrow$ negative work on striker (decreases striker's energy).
  • White coin hits black coin: White coin does positive work on black coin (increases black coin's energy); black coin does negative work on white coin (decreases white coin's energy).

Forms of Energy and Mechanical Energy

• Taxonomy of Energy Forms:
  • Mechanical Energy: Energy due to motion or spatial position of objects.
  • Thermal Energy: Kinetic energy of microscopic atomic/molecular motion making substances warm.
  • Light Energy: Electromagnetic radiation enabling vision.
  • Sound Energy: Energy carried by mechanical vibrations of air or medium molecules.
  • Electrical Energy: Energy associated with position or motion of electric charges.
  • Nuclear Energy: Binding energy stored within atomic nuclei.
  • Chemical Energy: Potential energy stored in chemical bonds between atoms in fuels and food.

• Energy Interconversions:

  • Light bulb: Electrical energy $\rightarrow$ Light energy.
  • Water heater: Electrical energy $\rightarrow$ Thermal energy.
  • Human muscle: Food chemical energy $\rightarrow$ Mechanical energy.
  • Ringing bell: Mechanical energy $\rightarrow$ Sound energy.

• Pause and Ponder Answer 3 (Bicycle Riding):

  • Pedaling converts muscular chemical energy into:
  1. Mechanical kinetic energy of the bicycle/rider motion.
  2. Thermal energy in muscles, tires, and bearing friction.
  3. Sound energy from tire-road interaction and chain mechanics.

Mechanical Energy: Kinetic Energy

• Definition:

  • Kinetic energy ($K$) is the energy possessed by an object due to its motion.
  • An object at rest has zero initial kinetic energy ($K = 0$).

• Derivation of Kinetic Energy Formula:

  • Consider an object of mass $m$ moving along a straight line with initial velocity $u$. A constant force $F$ produces uniform acceleration $a$ over displacement $s$, changing velocity to $v$.
  • From kinematic equation of motion:   $v^2 = u^2 + 2as$$s = \frac{v^2 - u^2}{2a}$
  • From definition of work $W = F \times s$ and Newton's second law $F = ma$:   $W = ma \times \frac{v^2 - u^2}{2a}$$W = \frac{1}{2} m (v^2 - u^2)$
  • If the object accelerates from rest ($u = 0$), the total work done equals its final kinetic energy $K$:   $K = \frac{1}{2} m v^2$

• Properties of Kinetic Energy:

  • SI unit: Joule ($\text{J}$).
  • Strictly scalar (independent of velocity direction).
  • Dependent on mass ($m$) linearly and velocity magnitude ($v$) quadratically.
  • Positive work increases $K$; negative work decreases $K$; zero work maintains constant $K$

• Worked Examples:

  • Example 7.4 (Velocity Doubling):
  • Initial kinetic energy: $K_1 = \frac{1}{2} m v^2$
  • Doubled velocity $v_2 = 2v$:     $K_2 = \frac{1}{2} m (2v)^2 = 4 \left(\frac{1}{2} m v^2\right) = 4 K_1$
  • Kinetic energy increases by a factor of 4.
  • Example 7.5 (Fast Cricket Ball):
  • Mass $m = 0.2\,\text{kg}$, velocity $v = 154.8\,\text{km\,h}^{-1} = \frac{154.8 \times 1000}{3600} = 43\,\text{m\,s}^{-1}$.     $K = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.2\,\text{kg} \times (43\,\text{m\,s}^{-1})^2 = 0.1 \times 1849 = 184.9\,\text{J}$
  • Example 7.6 (Jet Landing on Aircraft Carrier):
  • Mass $m = 15000\,\text{kg}$, backward force $F = 367500\,\text{N}$, distance $s = 100\,\text{m}$, final velocity $v_f = 0\,\text{m\,s}^{-1}$.
  • Work done by cable: $W = -F \times s = -367500\,\text{N} \times 100\,\text{m} = -36750000\,\text{J}$.
  • By Work-Energy Theorem: $\Delta K = W$$K_f - K_i = 0 - \frac{1}{2} m v^2 = -36750000$$\frac{1}{2} \times 15000\,\text{kg} \times v^2 = 36750000$$7500 v^2 = 36750000$$v^2 = 4900\,\text{m}^2\,\text{s}^{-2}$$v = 70\,\text{m\,s}^{-1} = 70 \times 3.6 = 252\,\text{km\,h}^{-1}$

• Pause and Ponder Answers:
  4. Ratio of velocities for equal kinetic energy: Objects A ($m_A = m$) and B ($m_B = 4m$) have $K_A = K_B$.      $\frac{1}{2} m v_A^2 = \frac{1}{2} (4m) v_B^2 \implies v_A^2 = 4 v_B^2 \implies \frac{v_A}{v_B} = 2$      The ratio of velocity magnitudes is $2:1$.
  5. Kinetic energy during constant velocity horizontal motion: No. Since mass and velocity are constant, kinetic energy remains constant regardless of spatial position.

Mechanical Energy: Potential Energy

• Conceptual Basis & Physical Manifestations:
  • Potential energy ($U$) is energy stored in a body or system due to its structural deformation, internal configuration, or relative spatial position against conservative force fields.
  • Deformation Examples:
  • Slingshot (Gulel): Stretching rubber bands stores energy. Upon release, stored elastic potential energy transfers into stone kinetic energy.
  • Bow and Arrow: Drawing the bowstring bends the wooden limbs, storing elastic strain energy that propels the arrow.
  • Spring: Compressing or extending a helical spring against internal restoring forces stores mechanical energy.

• Relative Spatial Configuration Examples:
  • Magnetic Poles: Pulling apart attractive opposite magnetic poles requires work against magnetic force, storing energy in the field. Releasing them allows conversion to kinetic energy.
  • Electric Charges: Work done separating opposite charges stores electrostatic potential energy.
  • Earth-Ball System: Work done raising a mass against gravity stores gravitational potential energy in the Earth-mass system.

• Gravitational Potential Energy Formulation:
  • Earth is vastly more massive than near-surface objects, so system stored energy is conventionally attributed solely to the object.
  • Activity 7.1 (Sand Bed Impact):
  • Dropping a ball from $1\,\text{m}$ creates a shallow depression.
  • Dropping the same ball from $2\,\text{m}$ creates a significantly deeper depression.
  • Greater release height requires more work to reach, storing higher potential energy that performs more work on the sand.

• Mathematical Derivation:
  • Setting reference ground level at $h = 0$ where $U = 0$.
  • To raise an object of mass $m$ slowly without acceleration to vertical height $h$, applied force $F = mg$.
  • Displacement $s = h$ vertically upwards.
  • Work done by lifting force:   $W = F \times s = mg \times h = mgh$
  • By Work-Energy Theorem, this work equals stored gravitational potential energy:   $U = mgh$

• Worked Examples & Limitations:
  • Example 7.7 (Cricket Ball Height):
  • Mass $m = 200\,\text{g} = 0.2\,\text{kg}$, height $h = 10\,\text{m}$, $g = 10\,\text{m\,s}^{-2}$.     $U = mgh = 0.2\,\text{kg} \times 10\,\text{m\,s}^{-2} \times 10\,\text{m} = 20\,\text{J}$
  • Pause and Ponder Answer 6:
  • Horizontal constant velocity motion does not alter height ($\Delta h = 0$), so potential energy remains unchanged.
  • Vertical elevation increases $h$, proportionally increasing potential energy.
  • Limitations & Exceptions:
  • Conservative vs Non-conservative: Potential energy storage occurs only when work is done against conservative forces (gravitational, electrostatic, magnetic). Work against friction produces heat and does not store potential energy.
  • Field Variation: Formula $U = mgh$ is valid strictly near Earth's surface where $g$ is constant. At large radial distances, $g$ decreases with distance.

Conservation of Mechanical Energy

• Total Mechanical Energy:

  • The sum of kinetic energy and potential energy is defined as mechanical energy:   $E_{\text{mech}} = K + U$

• Mathematical Proof of Conservation in Free Fall:

  • Consider a mass $m$ dropped from rest ($u = 0$) at point A at height $h$ above ground.

1. At Point A (Top, $t = 0$):

   • Potential Energy: $U_A = mgh$
   • Kinetic Energy: $K_A = 0$
   • Total Mechanical Energy: $E_{A} = mgh + 0 = mgh$

2. At Point B (Intermediate position at time $t$, height $h'$ above ground):

   • Vertical distance fallen: $h - h' = \frac{1}{2} g t^2 \implies h' = h - \frac{1}{2} g t^2$
   • Downward velocity acquired: $v = gt$
   • Potential Energy:      $U_B = mgh' = mg\left(h - \frac{1}{2} g t^2\right) = mgh - \frac{1}{2} m g^2 t^2$
   • Kinetic Energy:      $K_B = \frac{1}{2} m v^2 = \frac{1}{2} m (gt)^2 = \frac{1}{2} m g^2 t^2$
   • Total Mechanical Energy:      $E_{B} = U_B + K_B = \left(mgh - \frac{1}{2} m g^2 t^2\right) + \left(\frac{1}{2} m g^2 t^2\right) = mgh$

3. At Point C (Just prior to ground impact, height $h = 0$):

   • Height $h = 0 \implies U_C = 0$
   • Velocity $v^2 = u^2 + 2gh = 2gh$
   • Kinetic Energy: $K_C = \frac{1}{2} m (2gh) = mgh$
   • Total Mechanical Energy: $E_{C} = 0 + mgh = mgh$

• Principle Statement:

  • In an isolated system subject only to conservative forces, total mechanical energy remains constant throughout motion. Decreases in potential energy convert into equal increases in kinetic energy.

• Activity 7.2 (Simple Pendulum Oscillation):

  • Pendulum bob released from height $h$ at extreme point P ($K=0$, $U=mgh$).
  • At lowest point Q, potential energy converts to kinetic energy ($U=0$, $K_{\text{max}}$).
  • At opposite extreme point R, bob rises back to height $h$ ($K=0$, $U=mgh$).
  • Real pendulums slow down over time as mechanical energy dissipates into thermal energy via air resistance and pivot friction.

• Worked Examples:
  • Example 7.8 (Child on Slide):
  • Height $h$. Initial $U = mgh$, $K = 0$.
  • At bottom, $U = 0$, $K = \frac{1}{2} m v^2$$\frac{1}{2} m v^2 = mgh \implies v = \sqrt{2gh}$
  • Velocity depends solely on vertical drop $h$, independent of mass or slide slope shape (ignoring friction).
  • Example 7.9 (Truck Highway Escape Ramp):
  • Truck mass $m = 10000\,\text{kg}$, speed $v = 72\,\text{km\,h}^{-1} = 20\,\text{m\,s}^{-1}$, incline angle $30^\circ$ (height $h = \frac{d}{2}$ for ramp distance $d$), resistive sand force $F = 50000\,\text{N}$, $g = 10\,\text{m\,s}^{-2}$.
  • Initial total energy: $E_i = K_i + U_i = \frac{1}{2} (10000)(20)^2 + 0 = 2000000\,\text{J}$.
  • Final total energy at rest: $E_f = U_f = mgh = 10000 \times 10 \times \left(\frac{d}{2}\right) = 50000 d$
  • Work done by sand friction: $W_{\text{sand}} = -F \times d = -50000 d$
  • By Work-Energy Theorem:     $W_{\text{sand}} = E_f - E_i$$-50000 d = 50000 d - 2000000$$100000 d = 2000000$$d = 20\,\text{m}$
  • Minimum ramp length required is $20\,\text{m}$.

• Pause and Ponder Answers:
  7. Mechanical energy just before ground impact: At ground level $h=0$, potential energy $U = 0$. Velocity $v^2 = 2gh$, so $K = \frac{1}{2}m(2gh) = mgh$. Total mechanical energy $E = K + U = mgh + 0 = mgh$.\n 8. *Ball roller coaster in science park*: At A, U$ is max. As it moves to B, $U \rightarrow K$. At C, D, E, heights decrease progressively because friction and air resistance convert part of mechanical energy into heat and sound.\n\n\n\n# Power\n\n- **Definition**:\n - Power (P) is defined as the time rate at which work is done or energy is transferred.\n  P = \frac{W}{t}\n\n- **SI Unit and Equivalent Units**:\n - SI unit of power is the **watt** (\text{W}).\n  1\,\text{W} = 1\,\text{J\,s}^{-1}\n - One watt is the power of an agency that performs 1 joule of work per second.\n\n- **Horsepower (\text{hp})**:\n - Unit used in automotive and industrial machinery applications:\n  1\,\text{hp} = 746\,\text{W}\n - Originally established by James Watt to compare steam engine performance against draft horses.\n\n- **Meet a Scientist: James Watt**:\n - Invented the separate condenser steam engine capable of efficient rotational drive output.\n\n\n\n- **Worked Examples**:\n - **Example 7.10 (Weightlifter Power)**:\n - Mass m = 75\,\text{kg}$, height$h = 2\,\text{m}$, time$t = 5\,\text{s}$,$g = 10\,\text{m\,s}^{-2}.\n    W = mgh = 75\,\text{kg} \times 10\,\text{m\,s}^{-2} \times 2\,\text{m} = 1500\,\text{J}\n    P = \frac{1500\,\text{J}}{5\,\text{s}} = 300\,\text{W}\n - **Example 7.11 (Car Acceleration Power)**:\n - Mass m = 1000\,\text{kg}$,$u = 0$,$v = 72\,\text{km\,h}^{-1} = 20\,\text{m\,s}^{-1}$, time$t = 10\,\text{s}.\n    W = \Delta K = \frac{1}{2} m v^2 - 0 = \frac{1}{2} \times 1000\,\text{kg} \times (20\,\text{m\,s}^{-1})^2 = 200000\,\text{J}\n    P = \frac{200000\,\text{J}}{10\,\text{s}} = 20000\,\text{W} = 20\,\text{kW}\n\n# Simple Machines\n\n- **Fundamental Definitions**:\n - Devices that alter force magnitude or direction to make work easier without changing total required mechanical work.\n - **Effort**: Force applied to the machine.\n - **Load**: Resistance force to be overcome.\n - **Mechanical Advantage (MA)**:\n  \text{mechanical advantage} = \frac{\text{load}}{\text{effort}}\n\n- **Pulleys**:\n - **Fixed Pulley**: Grooved wheel mounted on a stationary axle. Re-routes effort direction (pulling downward with gravity is easier than lifting upward). \text{MA} = 1.\n - **Movable Pulley / Pulley Systems**: Load attaches to movable wheel frame. Effort applies to free rope end. \text{MA} > 1, allowing heavy loads to be lifted with reduced effort force.\n\n\n\n\n\n- **Inclined Plane**:\n - Sloped rigid surface used to elevate loads.\n - **Activity 7.3**: Spring balance cart pulled up ramps of varying slopes. Shallow slopes require lower effort forces over longer ramp lengths L.\n - **Derivation of Mechanical Advantage**:\n - Mass m$, load = weight$mg$. Height =$h$, ramp length =$L$, effort =$F'.\n - Work input W_{\text{in}} = F' \times L.\n - Potential energy gained U = mgh\n - Ignoring friction: F' \times L = mgh \implies \frac{mg}{F'} = \frac{L}{h}\n    \text{mechanical advantage} = \frac{\text{load}}{\text{effort}} = \frac{mg}{F'} = \frac{L}{h}\n - Since L > h$,$\text{MA} > 1.\n\n\n\n\n\n - **Worked Example 7.12 (Step Ramp)**:\n - Step height AB = 30\,\text{cm}$, base$BC = 40\,\text{cm}.\n - Ramp hypotenuse length AC = \sqrt{30^2 + 40^2} = 50\,\text{cm}.\n    \text{MA} = \frac{L}{h} = \frac{50\,\text{cm}}{30\,\text{cm}} = 1.67\n - **Pause and Ponder Answers**:\n 9. *Winding hill roads*: Switchback roads increase slope path distance L, reducing required vehicle tractive effort force.\n 10. *Inclined vs vertical ladders*: Inclined ladders act as inclined planes (L > h), requiring less muscular force than vertical climbs.\n\n- **Levers**:\n - Rigid bar pivoting around a fixed point called the **fulcrum**.\n - Key parameters: **Load Arm** (distance from fulcrum to load) and **Effort Arm** (distance from fulcrum to effort).\n - **Work Transfer Relation**:\n  F_1 \times d_1 = F_2 \times d_2\n - **Principle of Moments / Balance Equation**:\n  \text{effort} \times \text{effort arm} = \text{load} \times \text{load arm}\n  n_1 \times L_1 = n_2 \times L_2\n - **Mechanical Advantage of Lever**:\n  \text{mechanical advantage} = \frac{\text{load}}{\text{effort}} = \frac{\text{effort arm}}{\text{load arm}}\n\n\n\n\n\n- **Three Classes of Levers**:\n - **Class I**: Fulcrum located between Effort and Load. Examples: Tongs, scissors, crowbar, pliers, beam balance, seesaw.\n - **Class II**: Load located between Fulcrum and Effort. Examples: Lemon squeezer, wheelbarrow, bottle opener.\n - **Class III**: Effort located between Fulcrum and Load. Examples: Tweezers, broom, hammer, oar, forearm.\n\n- **Worked Example 7.13 (Seesaw Balance)**:\n - Seesaw seats A, B, C, D, E. AC = EC = 2\,\text{m}$,$BC = DC = 1\,\text{m}.\n - Child 1 (15\,\text{kg}$) sits at seat A ($L_1 = 2\,\text{m}$). Child 2 ($30\,\text{kg}$) sits at distance$L_2.\n  15\,\text{kg} \times 2\,\text{m} = 30\,\text{kg} \times L_2 \implies L_2 = 1\,\text{m}\n - Child 2 must sit at seat D.\n\n\n\n- **Pause and Ponder Answers**:\n 11. *Spoon opening can lid*: Acts as Class I lever with effort arm much longer than load arm, yielding high mechanical advantage.\n 12. *Cutting hard objects with scissors*: Placing object near fulcrum decreases load arm length, increasing mechanical advantage.\n 13. *Perpetual motion impossibility*: Real machines always experience mechanical friction and drag, steadily converting mechanical work into unrecoverable thermal energy.\n\n# Bridging Science and Society & Review Exercises\n\n- **Himalayan Watermill (*Gharat* or *Panchakki*)**:\n - Downhill flowing stream converts gravitational potential energy into kinetic energy.\n - Water streams through pipe A onto wooden turbine wheel B, generating rotational motion that turns heavy upper grinding stone C.\n - Modern hydroelectric dams operate on the exact same potential-to-kinetic conversion principle to drive generators.\n\n\n\n- **The Quest Continues: Dark Energy**:\n - Modern astronomical observations show universe expansion is accelerating, attributed to a mysterious energy form termed dark energy.\n\n- **Complete Solutions to Review Exercises**:\n 1. **True/False**:\n     (i) *False* (Work requires non-zero displacement).\n     (ii) *True* (Upward force aligns with upward displacement).\n     (iii) *True* (Both measured in joules).\n     (iv) *False* (Stretched band stores potential energy, not kinetic).\n     (v) *True* (Energy converts between forms).\n 2. **Fill in the Blanks**:\n     (i) Force \times Displacement.\n     (ii) 1 newton.\n     (iii) \frac{1}{2} m v^2.\n     (iv) mgh.\n     (v) Rate.\n 3. **Ball at Peak of Upward Throw**:\n - Correct options: **(iii) Its kinetic energy is zero**, and **(iv) Its potential energy is maximum**.\n 4. **Energy Transformations**:\n     (i) *Truck uphill*: Fuel chemical energy \rightarrow Kinetic & Gravitational potential energy.\n     (ii) *Watch spring unwinding*: Elastic potential energy \rightarrow Kinetic energy.\n     (iii) *Photosynthesis*: Solar light energy \rightarrow Chemical energy.\n     (iv) *Dam water release*: Gravitational potential energy \rightarrow Kinetic energy.\n     (v) *Burning matchstick*: Chemical energy \rightarrow Thermal & Light energy.\n     (vi) *Firecracker*: Chemical energy \rightarrow Kinetic, Thermal, Light, & Sound energy.\n     (vii) *Microphone*: Acoustic sound energy \rightarrow Electrical energy.\n     (viii) *Electric bulb*: Electrical energy \rightarrow Light & Thermal energy.\n     (ix) *Solar panel*: Light energy \rightarrow Electrical energy.\n 5. **Elevator vs Stairs Potential Energy** (m=50\,\text{kg}$,$h=72.5\,\text{m}$,$g=10\,\text{m\,s}^{-2}):\n     (i) Elevator: \Delta U = mgh = 50 \times 10 \times 72.5 = 36250\,\text{J}.\n     (ii) Stairs: \Delta U = mgh = 36250\,\text{J}.\n     (iii) Conclusion: Gravitational potential energy change depends strictly on initial and final elevation, completely independent of path taken.\n 6. **Crane Floor Heights**:\n - Raising same mass to 20th floor doubles elevation (2h$)$\rightarrow$Requires **double the energy** ($2E).\n - Double height in double time (2t$):$P = \frac{2E}{2t} = \frac{E}{t}$$\rightarrow Required **power remains identical**.\n 7. **Flag Raising**:\n - Work done (W = mgh) depends on flag weight and pole height; speed does not change work done.\n - Raising flag twice as fast halves time, **doubling the required power**.\n 8. **Scooter Fuel Comparison**:\n - Day 1 mass m_1 = 60 + 100 = 160\,\text{kg}$. Day 2 mass$m_2 = 60 + 100 + 40 = 200\,\text{kg}.\n - Kinetic energy ratio: \frac{K_2}{K_1} = \frac{\frac{1}{2} (200) v^2}{\frac{1}{2} (160) v^2} = \frac{200}{160} = \frac{5}{4} = 1.25.\n - Fuel consumption ratio is **5:4** (or 1.25).\n 9. **Seesaw Adult/Child Balance**:\n - Adult mass 2m$, Child mass$m$. Balance requires$m \times L_{\text{child}} = 2m \times L_{\text{adult}}.\n - Adult must sit at **half the distance** of the child relative to the fulcrum.\n 10. **Upward Ball Throw** (m=2\,\text{kg}$,$u=20\,\text{m\,s}^{-1}$, max height$h=19.4\,\text{m}):\n      (i) Gravity work sign: **Negative** during ascent; **Positive** during descent.\n      (ii) Initial K_i = \frac{1}{2} (2) (20)^2 = 400\,\text{J}$. Actual peak$U = mgh = 2 \times 10 \times 19.4 = 388\,\text{J}$. Work by air resistance:$388 - 400 = -12\,\text{J}.\n 11. **Variable Force Block** (m=10\,\text{kg}$, initial$K_0 = 180\,\text{J}):\n - Area under triangular force graph (0$to$4\,\text{m}$):$\text{Area} = \frac{1}{2} \times 4\,\text{m} \times 50\,\text{N} = 100\,\text{J}.\n      (i) Speed at 0\,\text{m}$:$\frac{1}{2} (10) v_0^2 = 180 \implies 5 v_0^2 = 180 \implies v_0 = 6\,\text{m\,s}^{-1}.\n      (ii) Speed at 4\,\text{m}$:$K_4 = 180 + 100 = 280\,\text{J} \implies 5 v_4^2 = 280 \implies v_4 = \sqrt{56} \approx 7.48\,\text{m\,s}^{-1}.\n - No, acceleration is never negative because force remains positive throughout.\n 12. **Lunar Vertical Throw**:\n - Lunar surface gravity g_{\text{moon}} = \frac{1}{6} g_{\text{earth}}.\n - Max throw height formula h = \frac{u^2}{2g}.\n - On Moon: h_{\text{moon}} = 6 \times h_{\text{earth}} = 6 \times 8\,\text{m} = 48\,\text{m}.\n 13. **Car Braking Dynamics** (m=1000\,\text{kg}$, constant speed$v=35\,\text{m\,s}^{-1}):\n      (i) Position A to B: Car travels at uniform velocity 35\,\text{m\,s}^{-1} during driver reaction time.\n      (ii) Kinetic energy at A: K = \frac{1}{2} (1000) (35)^2 = 500 \times 1225 = 612500\,\text{J}.\n      (iii) Work done by brakes (B to C): -612500\,\text{J}.\n      (iv) Kinetic energy transforms into thermal friction energy and acoustic sound energy.\n 14. **Potential Energy Graph Ball** (m=0.5\,\text{kg}$, total mechanical energy$E=30\,\text{J}):\n - Velocity formula: K = E - U = 30 - U \implies \frac{1}{2} (0.5) v^2 = 30 - U \implies 0.25 v^2 = 30 - U\n - At P (U = 10\,\text{J}$):$0.25 v_P^2 = 20 \implies v_P = \sqrt{80} \approx 8.94\,\text{m\,s}^{-1}.\n - At Q (U = 20\,\text{J}$):$0.25 v_Q^2 = 10 \implies v_Q = \sqrt{40} \approx 6.32\,\text{m\,s}^{-1}.\n - At R (U = 0\,\text{J}$):$0.25 v_R^2 = 30 \implies v_R = \sqrt{120} \approx 10.95\,\text{m\,s}^{-1}.\n 15. **Falling Coconut** (m=1.5\,\text{kg}$, height$h=10\,\text{m}$, resistive sand force$F=3000\,\text{N}$,$g=10\,\text{m\,s}^{-2}):\n      (i) Impact velocity: v = \sqrt{2gh} = \sqrt{2 \times 10 \times 10} = \sqrt{200} \approx 14.14\,\text{m\,s}^{-1}.\n      (ii) Energy at impact E = mgh = 1.5 \times 10 \times 10 = 150\,\text{J}.\n - Work done stopping coconut: F \times d = 150 \implies 3000 d = 150 \implies d = 0.05\,\text{m} = 5\,\text{cm}$$.