Comprehensive Study Guide: Work, Energy, Power, and Energy Conservation

Work Done (Kerja)

• Definition of Work Done (Kerja / Kerja yang Dilakukan):

  • Work done is defined as the product of the applied force ($F$) and the displacement ($s$) of an object in the direction of the force.

  • Mathematical formula: $W = F \times s$

  • $W$ represents Work Done in Joules ($J$) or Newton-meters ($N\,m$).

  • $F$ represents Force in Newtons ($N$).

  • $s$ represents Displacement in meters ($m$).

• Displacement (Sesaran) vs. Distance (Jarak):

  • Displacement ($s$) is defined as the straight-line distance traveled in a specific, specified direction.

  • Distance ($s$) is the total path length traveled regardless of direction.

  • For general Form 3 calculations, displacement is treated as equal to distance ($\text{displacement} = \text{distance}$).

• Units of Measurement for Work:

  • Standard SI Unit: Joule ($J$) or Newton-meter ($N\,m$).

  • Conversion equivalence: $1\,J = 1\,N\,m = 1\,N \times 1\,m$.

  • One Joule ($1\,J$) of work is performed when a force of one Newton ($1\,N$) moves an object over a displacement of one meter ($1\,m$) in the direction of the force.

  • Larger SI prefix units:

  • Kilojoule ($kJ$): $1\,kJ = 1000\,J$

  • Megajoule ($MJ$): $1\,MJ = 10^6\,J = 1000000\,J$

• Conditions for Work to be Performed:

  • A force must act upon an object.

  • The object must undergo displacement ($s > 0$).

  • The displacement must be in the same direction as the applied force.

  • If displacement is zero ($s = 0$), the work done is exactly zero ($W = 0$).

• Examples and Hypotheses of Zero Work Done ($W = 0$):

  • Pushing against a solid stationary wall: Even if a force of $20\,N$ is exerted, the wall does not move ($s = 0\,m$). Therefore, $W = 20\,N \times 0\,m = 0\,J$.

  • A Royal Guard standing completely motionless on duty at Buckingham Palace for a 2-hour shift: Because displacement $s = 0\,m$, no scientific work is performed ($W = 0\,J$).

  • Watching television sitting stationary on a couch: Displacement $s = 0\,m$, hence $W = 0\,J$.

  • Sleeping on a bed or holding a stationary object without moving it vertically or horizontally: $s = 0\,m$, hence $W = 0\,J$.

• Working Against Gravity vs. With Gravity:

  • Lifting an object vertically upwards requires applying a force equal to the object's weight against the force of gravity ($g = 10\,m/s^2$ or $1\,kg = 10\,N$). Scientific work is performed by the person.

  • Dropping an object downwards allows gravity to do the work; the person holding/releasing it does no work during the fall.

  • Climbing stairs requires carrying full body weight vertically upwards against gravity. Therefore, only the vertical height is used as displacement ($s$) in calculating work done against gravity.

• Step-by-Step Worked Examples of Work Done:

  • Example 1 (Student carrying load up stairs):

  • A student weighing $400\,N$ carries a backpack load of $100\,N$ up stairs with a vertical height of $3\,m$.

  • Total force $F = 400\,N + 100\,N = 500\,N$.

  • Work done $W = F \times s = 500\,N \times 3\,m = 1500\,J$.

  • Example 2 (Worker pulling cement via pulley):

  • A worker uses a pulley system (puli) to pull a bucket of cement weighing $300\,N$ to the 1st floor at a height of $10\,m$.

  • Work done $W = F \times s = 300\,N \times 10\,m = 3000\,J$.

  • Example 3 (Amran lifting mass with pulley):

  • Amran pulls a box of mass $m = 15\,kg$ ($F = 150\,N$) over a vertical height of $4\,m$.

  • Work done $W = 150\,N \times 4\,m = 600\,J$.

  • Example 4 (Pushing a trolley horizontally):

  • A force of $10\,N$ pushes a shopping trolley horizontally for $5\,m$.

  • Work done $W = 10\,N \times 5\,m = 50\,J$.

  • Example 5 (Alice lifting a box onto a cabinet):

  • Alice lifts a box of mass $m = 10\,kg$ ($F = 100\,N$) vertically from the ground to a height of $2\,m$ on top of a cabinet (almari).

  • Work done $W = 100\,N \times 2\,m = 200\,J$.

  • Example 6 (Vincent pulling a wooden block):

  • Vincent pulls a wooden block horizontally across the floor using a force of $2\,N$ for $0.5\,m$.

  • Work done $W = 2\,N \times 0.5\,m = 1\,J$.

  • Example 7 (Pulling open a drawer):

  • A drawer is pulled open with a force of $2\,N$ over a displacement of $30\,cm$ ($s = 0.3\,m$).

  • Work done $W = 2\,N \times 0.3\,m = 0.6\,J$.

  • Example 8 (Unrolling toilet paper and unrolling a carpet):

  • Unrolling toilet paper by hand or unrolling a carpet performs work because force is applied and the material undergoes displacement in the direction of the force.

  • Example 9 (Shot Put Shot Comparison):

  • Student A extends hand $1.1\,m$; Student B extends hand fully $1.2\,m$. Student B does more work because the displacement of the iron ball in the direction of force is greater.

  • For an iron ball experiencing a force of $20\,N$ over a displacement of $0.5\,m$: $W = 20\,N \times 0.5\,m = 10\,J$.

Power (Kuasa)

• Definition of Power (Kuasa):

  • Power is defined as the rate of work done (kadar kerja yang dilakukan) or the rate of energy transferred/consumed per unit time.

  • Mathematical formula: $P = \frac{W}{t} = \frac{F \times s}{t}$

  • $P$ represents Power in Watts ($W$) or Joules per second ($J/s$).

  • $W$ represents Work Done in Joules ($J$).

  • $t$ represents Time taken in seconds ($s$).

• Units of Measurement for Power:

  • Standard SI Unit: Watt ($W$) or Joule per second ($J/s$).

  • Conversion equivalence: $1\,W = 1\,J/s$.

  • Prefix units:

  • Kilowatt ($kW$): $1\,kW = 1000\,W$

  • Megawatt ($MW$): $1\,MW = 10^6\,W = 1000000\,W$

• Conceptual Distinction: Energy vs. Power (The Ferrari vs. Myvi Analogy):

  • Both a Perodua Myvi and a Ferrari require the same work/energy ($10000\,J$) to accelerate to $100\,km/h$.

  • A Myvi takes $10\,s$ to do this work ($P = \frac{10000\,J}{10\,s} = 1000\,W$).

  • A Ferrari takes $1\,s$ to do this work ($P = \frac{10000\,J}{1\,s} = 10000\,W$).

  • The Ferrari is more expensive and powerful because it performs the same work in a significantly shorter time frame.

• Real-World Electrical Power Ratings:

  • Air Conditioner: Rated at approximately $750\,W$, meaning it consumes $750\,J$ of electrical energy per second.

  • LED Light Bulb: Rated at approximately $10\,W$, meaning it consumes $10\,J$ of electrical energy per second.

  • Household Electricity Meters: Measure energy consumption in Kilowatt-hours ($kWh$).

• Step-by-Step Worked Examples of Power Calculations:

  • Example 1 (Monkey climbing a tree):

  • A monkey weighing $50\,N$ climbs an tree to a height of $3\,m$ in $20\,s$.

  • Work done $W = 50\,N \times 3\,m = 150\,J$.

  • Power P = \frac{150\,J}{20\,s} = 7.5\,W$.\n - Example 2 (Monkey pulling a box):\n - A monkey pulls a box with F = 30\,N$over$s = 2\,m$in$t = 5\,s.\n - Work done W = 30\,N \times 2\,m = 60\,J.\n - Power P = \frac{60\,J}{5\,s} = 12\,W$.

  • Example 3 (Weightlifter):

  • A weight of $150\,N$ is lifted to a height of $1\,m$ in $0.5\,s$.

  • Work done $W = 150\,N \times 1\,m = 150\,J$.

  • Power P = \frac{150\,J}{0.5\,s} = 300\,W$.\n - Example 4 (Sprinting Athlete):\n - An athlete with mass m = 30\,kg$($F = 300\,N$) runs a$120\,m$distance in$20\,s.\n - Work done W = 300\,N \times 120\,m = 36000\,J.\n - Power P = \frac{36000\,J}{20\,s} = 1800\,W$.

  • Example 5 (Motor vehicle acceleration):

  • A car weighing $20000\,N$ travels $1\,km$ ($1000\,m$) in $300\,s$ ($5\,\text{minutes}$).

  • Work done W = 20000\,N \times 1000\,m = 20000000\,J$.\n - Power P = \frac{20000000\,J}{300\,s} = 66666.67\,W = 66.67\,kW$.

  • Example 6 (Construction Crane):

  • A crane lifts a load of $1000\,N$ to a height of $150\,m$ in $10\,s$.

  • Power P = \frac{1000\,N \times 150\,m}{10\,s} = 15000\,W = 15\,kW$.\n - Example 7 (Boy climbing stairs):\n - A boy of mass m = 50\,kg$($F = 500\,N$) climbs stairs with a vertical height of$2\,m$in$5\,s.\n - Power P = \frac{500\,N \times 2\,m}{5\,s} = 200\,W$.

  • Example 8 (Trainees running uphill):

  • Surface inclined height $h = 3\,m$.

  • Trainee P ($m = 48\,kg \rightarrow 480\,N$), mean time $t = 5.5\,s$: Power P = \frac{480\,N \times 3\,m}{5.5\,s} = 261.82\,W$.\n - Trainee Q (m = 52\,kg \rightarrow 520\,N$), mean time$t = 6.1\,s$: Power$P = \frac{520\,N \times 3\,m}{6.1\,s} = 255.74\,W$.

  • Trainee R ($m = 49\,kg \rightarrow 490\,N$), mean time $t = 6.13\,s$: Power P = \frac{490\,N \times 3\,m}{6.13\,s} = 239.80\,W$.\n - Analysis: Trainee P completes the run in the shortest time, but Trainee Q generates the maximum mechanical power due to higher mass.\n - Example 9 (Lifting weights to table height h = 1.5\,m):\n - Weight P (15\,kg \rightarrow 150\,N$) lifted in$10\,s$: Power$P = \frac{150 \times 1.5}{10} = 22.5\,W$.

  • Weight Q ($20\,kg \rightarrow 200\,N$) lifted in $15\,s$: Power P = \frac{200 \times 1.5}{15} = 20\,W$.\n - Weight R (25\,kg \rightarrow 250\,N$) lifted in$20\,s$: Power$P = \frac{250 \times 1.5}{20} = 18.75\,W$.

  • Principle derived: Power output is inversely proportional to the time taken to perform a given work quantity.

Gravitational Potential Energy (Tenaga Keupayaan Graviti - TKG)

• Definition of Gravitational Potential Energy (GPE / TKG):

  • GPE is defined as the work done to lift an object to a height ($h$) against gravitational acceleration.

  • Mathematical formula: $GPE = mgh$

  • $m$ represents Mass in kilograms ($kg$).

  • $g$ represents Gravitational Acceleration in meters per second squared ($m/s^2$) or Newtons per kilogram ($N/kg$). On Earth, $g = 10\,m/s^2$ (standard fixed value for Form 3 calculations).

  • $h$ represents Vertical Height in meters ($m$).

  • Unit: Joule ($J$).

• Factors Affecting GPE:

  • Mass ($m$): GPE is directly proportional to mass. A heavier object at the same height stores more energy than a lighter object.

  • Height ($h$): GPE is directly proportional to height above the reference ground level. The higher an object is lifted, the more energy it stores.

• Comparative Mass Scenarios (Ifat vs. Sam):

  • Ifat ($m = 40\,kg$) and Sam ($m = 55\,kg$) climb stairs of height $h = 2\,m$:

  • Ifat: $GPE = 40\,kg \times 10\,m/s^2 \times 2\,m = 800\,J$

  • Sam: $GPE = 55\,kg \times 10\,m/s^2 \times 2\,m = 1100\,J$

  • Conclusion: Sam stores and expends more energy because he has a larger mass. This is analogous to a heavy lorry consuming more fuel to move than a lightweight passenger car.

• Step-by-Step Worked Examples of GPE:

  • Example 1 (Lifting a box):

  • Mass $m = 3\,kg$, height $h = 2\,m$, $g = 10\,m/s^2$

  • $GPE = 3\,kg \times 10\,m/s^2 \times 2\,m = 60\,J$

  • Interpretation: The muscle performs $60\,J$ of work, transferring $60\,J$ of energy to be stored inside the box as GPE.

  • Example 2 (High-jump athlete):

  • Mass $m = 50\,kg$, height $h = 5\,m$, $g = 10\,m/s^2$

  • $GPE = 50\,kg \times 10\,m/s^2 \times 5\,m = 2500\,J$

  • Example 3 (Reverse Calculation - Mass determination):

  • Object placed at height $h = 8\,m$ possesses $GPE = 2000\,J$.

  • $2000 = m \times 10 \times 8 \Rightarrow 2000 = 80m \Rightarrow m = \frac{2000}{80} = 25\,kg$

  • Example 4 (Passenger Lift / Elevator):

  • Lift carries maximum load mass $m = 1500\,kg$ (approx. 20+ people at $70\,kg$ average) to height h = 30\,m$.\n - GPE = 1500\,kg \times 10\,m/s^2 \times 30\,m = 450000\,J\n - Relationship to Work Done: The electrical work done by the lift motor is equal to the GPE gained by the lift payload (W = GPE = 450000\,J).\n - Example 5 (Metal ball on a stand):\n - Mass m = 1.5\,kg$, height$h = 75\,cm = 0.75\,m.\n - GPE = 1.5\,kg \times 10\,m/s^2 \times 0.75\,m = 11.25\,J\n - Example 6 (Watermelon on Table - Earth vs. Moon comparison):\n - On Earth: Watermelon mass m = 5\,kg$, table height$h = 0.8\,m$,$g = 10\,m/s^2.\n - GPE = 5\,kg \times 10\,m/s^2 \times 0.8\,m = 40\,J\n - On the Moon: Gravitational acceleration g_{\text{Moon}} = \frac{1}{6} g_{\text{Earth}} \approx 1.67\,m/s^2$.

    • $GPE_{\text{Moon}} = 5\,kg \times 1.67\,m/s^2 \times 0.8\,m = 6.68\,J$

  • Phenomenon Explanation: When dropped, the watermelon on Earth breaks because it releases $40\,J$ of kinetic energy onto the floor. On the Moon, because $g$ is lower, it stores substantially less GPE ($6.68\,J$), releasing insufficient kinetic energy upon impact to break the watermelon shell.

Elastic Potential Energy (Tenaga Keupayaan Kenyal - TKE)

• Definition of Elastic Potential Energy (EPE / TKE):

  • EPE is defined as the work done to compress or stretch an elastic material (such as a spring or rubber band) from its original equilibrium position.

  • Mathematical formula: $EPE = \frac{1}{2} F x$

  • $F$ represents Force applied in Newtons ($N$).

  • $x$ represents Displacement/Extension/Compression from equilibrium length in meters ($m$).

  • Graphical Interpretation: EPE equals the area under the Force versus Extension ($F-x$) graph.

  • Unit: Joule ($J$).

• Step-by-Step Worked Examples of EPE Calculations:

  • Example 1 (Compressed spring):

  • A force $F = 5\,N$ compresses a spring from original length $50\,cm$ to $25\,cm$.

  • Extension/Compression x = 50\,cm - 25\,cm = 25\,cm = 0.25\,m$.\n - EPE = \frac{1}{2} \times 5\,N \times 0.25\,m = 0.625\,J\n - Example 2 (Elongated spring):\n - A spring is stretched with force F = 60\,N$producing extension$x = 20\,cm = 0.2\,m$.

  • $EPE = \frac{1}{2} \times 60\,N \times 0.2\,m = 6\,J$

  • Physical effect: Releasing this rubber band onto skin causes pain because $6\,J$ of stored energy is abruptly transferred to the skin.

  • Example 3 (Spring compressed by a mass weight):

  • Original spring length = $20\,cm = 0.2\,m$. Compressed with mass $m = 2\,kg$ ($F = 20\,N$) until length decreases to 12\,cm = 0.12\,m$.\n - Displacement x = 0.2\,m - 0.12\,m = 0.08\,m$.

  • $EPE = \frac{1}{2} \times 20\,N \times 0.08\,m = 0.8\,J$

  • Example 4 (Bow and Arrow / Catapult string):

  • A bow string is pulled with a force $F = 25\,N$ over displacement x = 0.5\,m$.\n - EPE = \frac{1}{2} \times 25\,N \times 0.5\,m = 6.25\,J\n\n# Kinetic Energy (*Tenaga Kinetic* - TK)\n\n- Definition of Kinetic Energy (KE / *TK*):\n - Kinetic Energy is defined as the energy possessed by a moving object (*objek yang bergerak*).\n - Mathematical formula: KE = \frac{1}{2} m v^2\n - m$represents Mass in kilograms ($kg).\n - v$represents Velocity or Speed (*halaju* / *kelajuan*) in meters per second ($m/s$or$m\,s^{-1}).\n - Unit: Joule (J).\n\n- Key Characteristics:\n - Any stationary object (v = 0\,m/s$) has zero kinetic energy ($KE = 0\,J).\n - Kinetic energy increases quadratically with velocity (KE \propto v^2).\n\n- Step-by-Step Worked Examples of KE Calculations:\n - Example 1 (High-speed train):\n - Train mass m = 500000\,kg$, velocity converted to SI unit$v = 1000\,m/s.\n - KE = \frac{1}{2} \times 500000\,kg \times (1000\,m/s)^2 = 2.5 \times 10^{11}\,J\n - Example 2 (Flying Superhero):\n - Superhero mass m = 50\,kg$, flying speed$v = 800\,m/s$.

  • $KE = \frac{1}{2} \times 50\,kg \times (800\,m/s)^2 = 1.6 \times 10^7\,J = 16000000\,J$

  • Example 3 (Ball Bearing - Finding velocity):

  • Ball bearing mass $m = 0.2\,kg$ has $KE = 3.6\,J$. Find velocity v$.\n - 3.6 = \frac{1}{2} \times 0.2 \times v^2 \Rightarrow 3.6 = 0.1 v^2 \Rightarrow v^2 = 36 \Rightarrow v = \sqrt{36} = 6\,m/s\n - Example 4 (Subatomic Electron):\n - Electron mass m = 9 \times 10^{-31}\,kg$, speed$v = 4 \times 10^6\,m/s$.

  • $KE = \frac{1}{2} \times (9 \times 10^{-31}\,kg) \times (4 \times 10^6\,m/s)^2 = 7.2 \times 10^{-18}\,J$

  • Example 5 (Jumping Flea / Kutu):

  • Flea mass $m = 0.5\,mg = 0.5 \times 10^{-3}\,g = 0.5 \times 10^{-6}\,kg$ (or $0.5 \times 10^{-3}\,kg$ in contextual SI practice), jumping speed v = 1\,m/s$.\n - KE = \frac{1}{2} \times (0.5 \times 10^{-3}) \times 1^2 = 0.25 \times 10^{-3}\,J = 0.00025\,J\n\n# Principle of Conservation of Energy (*Prinsip Keabadian Tenaga*)\n\n- Fundamental Law:\n - The Principle of Conservation of Energy states that energy cannot be created or destroyed (*tenaga tidak boleh dicipta atau dimusnahkan*).\n - Energy can only be transformed or converted from one form to another (e.g., Chemical Energy \rightarrow$GPE$\rightarrow$KE$\rightarrow Heat/Sound).\n - The total energy in an isolated system remains strictly constant.\n\n- Energy Flow and Conversion Chains:\n - Solar Radiation (Light Energy from Sun) \rightarrow$Photosynthesis in Plants (Chemical Energy / *Tenaga Kimia*)$\rightarrow$Animal Consumption (Chicken ingests plant chemical energy)$\rightarrow$Human Nutrition (Human eats chicken, gaining chemical energy)$\rightarrow$Muscle Contraction (Mechanical Work)$\rightarrow Gravitational Potential Energy / Elastic Potential Energy.\n\n- Real-World Implications of Energy Conservation:\n - Motor Vehicle Braking: Pressing a car brake pedal does not destroy the car's kinetic energy. It uses friction between brake pads and tire brake rotors to convert kinetic energy entirely into thermal/heat energy (*tenaga haba*), causing rotors to become sizzling hot.\n - High-Speed Crashes: In severe automobile collisions, excessive kinetic energy is instantly converted into sound energy (a loud bang), heat energy, and severe structural deformation. High energy transfer to passengers causes trauma or fatal injuries.\n - Automotive Safety Features: Seatbelts and airbags act to extend the impact distance and time, safely dissipating and transferring kinetic energy away from human internal organs.\n\n- Biological Energy Conservation and Weight Loss Science:\n - Human body operates as a thermodynamic machine fueled by chemical energy stored in food, measured in Joules (J$) or Calories ($kcal).\n - Daily Energy Requirement (Maintenance Level): E.g., a 66\,kg$human requires approximately$2200\,kcal/day to run basal metabolic functions.\n - Caloric Surplus: Eating 3000\,kcal$/day creates an excess of$+800\,kcal$. Because energy cannot be destroyed, the body converts and stores this extra$800\,kcal as subcutaneous adipose tissue (fat stores / body powerbank).\n - Caloric Deficit: Consuming 2000\,kcal$/day creates a deficit of$-200\,kcal$. The body extracts$200\,kcal from stored fat powerbanks, leading to weight loss.\n - Scientific Fact: Magic weight-loss diet pills claiming to destroy energy without caloric deficit violate the Principle of Conservation of Energy and are scientifically impossible.\n\n# Energy Efficiency and Useful vs. Wasted Energy\n\n- Useful Energy vs. Wasted Energy:\n - Useful Energy: Energy output converted into the intended, functional form to do desirable work.\n - Wasted Energy (*Tenaga Terbuang*): Energy output converted into unwanted forms (primarily friction heat and ambient noise).\n\n- System Efficiency Calculations & Comparisons:\n - Internal Combustion Engine Cars:\n - Efficiency: Only \approx 20\% useful energy (converted into KE).\n - Waste: \approx 80\% wasted energy lost as heat and noise.\n - Environmental impact: Millions of internal combustion engines dissipating 80\% thermal waste heat into the environment directly drives atmospheric global warming.\n - Electric Vehicles (e.g., Tesla):\n - Efficiency: \approx 80\% useful energy efficiency.\n - Waste: Only \approx 20\% wasted as heat.\n - Evolution of Lighting Efficiency:\n - Filament/Incandescent Bulbs: 8\%$efficient;$92\% wasted as extreme heat.\n - Fluorescent Lamps: 20\% - 30\%$efficient;$70\% - 80\% wasted as heat.\n - Modern LED Lamps: \approx 90\%$efficient; only$10\% wasted as heat.\n\n# Energy Transformations in Oscillating Systems\n\n- Definition of Oscillating Systems (*Sistem Ayunan*):\n - An oscillating system undergoes continuous, repetitive, to-and-fro motion about an equilibrium position in a fixed periodic pattern.\n\n- Case 1: Simple Pendulum (*Ayunan Liontin Ringkas*):\n - Motion Cycle Analysis (Positions X, Y, Z):\n - Position X (Highest extreme point on left):\n - Height h$is maximum$\rightarrow$$GPE = \text{Maximum}.\n - Pendulum bob stops instantaneously (v = 0\,m/s$)$\rightarrow$$KE = 0\,J.\n - Motion from Position X to Y (Swinging downwards):\n - Height decreases \rightarrow GPE decreases.\n - Speed increases \rightarrow KE increases.\n - Energy Transformation: GPE is converted into KE (GPE \rightarrow KE).\n - Position Y (Lowest equilibrium point):\n - Height h = 0$relative to baseline$\rightarrow$$GPE = 0\,J (Minimum).\n - Speed v$is fastest$\rightarrow$$KE = \text{Maximum}.\n - Motion from Position Y to Z (Swinging upwards):\n - Height increases \rightarrow GPE increases.\n - Speed decreases \rightarrow KE decreases.\n - Energy Transformation: KE is converted back into GPE (KE \rightarrow GPE).\n - Position Z (Highest extreme point on right):\n - Height h$is maximum$\rightarrow$$GPE = \text{Maximum}.\n - Bob stops instantaneously (v = 0\,m/s$)$\rightarrow$$KE = 0\,J.\n - Ideal Closed System vs. Real-Life System:\n - Closed System Ideal: Zero friction/air resistance. Energy transfers between GPE and KE with 100\%$efficiency ($GPE_{\text{max}} = KE_{\text{max}}). The pendulum swings indefinitely.\n - Real-Life System: Air resistance and pivot friction continuously convert mechanical energy into heat and sound energy lost to surrounding air. Swing amplitude gradually diminishes until motion stops.\n - The Bowling Ball Pendulum Demonstration: Releasing a heavy 10\,kg bowling ball pendulum from touching a person's nose will swing away and return *without* hitting the nose, because air resistance removes mechanical energy during oscillation.\n\n- Case 2: Loaded Spring System (*Sistem Beban Musang/Kenyal*):\n - Position X (Maximum extension/compression at bottom):\n - Extension x$is maximum$\rightarrow$$EPE = \text{Maximum}.\n - Mass block stops moving (v = 0\,m/s$)$\rightarrow$$KE = 0\,J.\n - Motion from Position X to Y (Moving towards equilibrium):\n - Spring extension shortens \rightarrow EPE decreases.\n - Block accelerates \rightarrow KE increases.\n - Energy Transformation: EPE converts to KE (EPE \rightarrow KE).\n - Position Y (Equilibrium position):\n - Displacement x = 0$$\rightarrow$$EPE = 0\,J.\n - Velocity v$is maximum$\rightarrow$$KE = \text{Maximum}.\n\n# Problem-Solving & Energy Calculations in Closed Systems\n\n- Definition of a Closed System:\n - A closed system is a system free from non-conservative external forces such as air resistance or surface friction. Total mechanical energy (E_{\text{total}} = PE + KE) is strictly conserved.\n\n- Problem 1: Compressed Toy Spring Pistol (EPE \rightarrow KE Conversion):\n - Scenario: Toy pistol spring original length = 300\,mm$. Compressed with force$F = 5\,N$to new length$50\,mm$. Plastic ball mass$m = 50\,g = 0.05\,kg.\n - Compression distance x = 300\,mm - 50\,mm = 250\,mm = 0.25\,m$.

  • Assumption: Closed system with zero friction loss ($EPE = KE$).

  • Step-by-Step Calculation:

  • $\frac{1}{2} F x = \frac{1}{2} m v^2$

  • $F x = m v^2$

  • $5\,N \times 0.25\,m = 0.05\,kg \times v^2$

  • $1.25 = 0.05 v^2$

  • $v^2 = \frac{1.25}{0.05} = 25$

  • $v = \sqrt{25} = 5\,m/s$

• Problem 2: Falling Durian from Tree (GPE $\rightarrow$ KE Conversion):

  • Scenario: A durian drops from a tree height $h = 5\,m$. Calculate impact velocity $v$ just before touching the ground ($g = 10\,m/s^2$).

  • Assumption: Closed system ($GPE = KE$).

  • Step-by-Step Calculation:

  • $mgh = \frac{1}{2} m v^2$

  • Mass $m$ cancels out from both sides: $gh = \frac{1}{2} v^2$

  • $v^2 = 2gh = 2 \times 10\,m/s^2 \times 5\,m = 100$

  • $v = \sqrt{100} = 10\,m/s$

• Problem 3: Vertical Catapult Shot (EPE $\rightarrow$ GPE Conversion):

  • Scenario: A stone mass $m = 20\,g = 0.02\,kg$ is loaded onto a catapult string with stored $EPE = 0.5\,J$. Shot $90^\circ$ vertically upwards. Calculate maximum height $h$.

  • Assumption: Closed system ($EPE = GPE$).

  • Step-by-Step Calculation:

  • $EPE = mgh$

  • $0.5\,J = 0.02\,kg \times 10\,m/s^2 \times h$

  • $0.5 = 0.2 h$

  • $h = \frac{0.5}{0.2} = 2.5\,m$

• Problem 4: Hypothetical High-Rise Fall (30th Floor):

  • Scenario: Jump from 30th floor height ($30 \times 3\,m = 90\,m$) under Earth gravity ($g = 10\,m/s^2$).

  • Calculation:

  • $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 90} = \sqrt{1800} \approx 42.43\,m/s$

  • Result: High impact speed produces catastrophic kinetic energy transfer upon contact, causing immediate fatal trauma.