Comprehensive Study Guide on Gravitational Potential Energy, Kinetic Energy, and Conservation of Mechanical Energy

Gravitational Potential Energy

Definition: The energy an object possesses due to its position above a reference point (typically the ground).
Fundamental Equation:
- $E_p = m \times g \times h$
- $E_p = mgh$
Variables, Standard Units, and Values:
- $E_p$ : Potential energy, measured in Joules ( $\text{J}$ ).
- $m$ : Mass of the object, which must strictly be in kilograms ( $\text{kg}$ ).
- $g$ : Gravitational acceleration (gravity). On Earth, $g = 9,8\,\text{m/s}^2$ .
- $h$ : Height of the object above the reference point, which must strictly be in metres ( $\text{m}$ ).
Boundary Condition for Height:
- If $h = 0\,\text{m}$ , then $E_p = 0\,\text{J}$ .
Proportionality and Relationships:
- As the mass ( $m$ ) of an object increases, its potential energy ( $E_p$ ) increases proportionally.
- As the height ( $h$ ) of an object increases, its potential energy ( $E_p$ ) increases proportionally.
Formula Triangle Representation:
- Top vertex: $E_p$
- Bottom divisions: $m$ , $g$ , $h$
- Rearrangements derived from the triangle:
- $m = \frac{E_p}{g \times h}$
- $h = \frac{E_p}{m \times g}$
- $g = \frac{E_p}{m \times h}$
Unit Conversion Rules for Potential Energy Calculations:
- Mass conversions:
- Convert grams ( $\text{g}$ ) to kilograms ( $\text{kg}$ ) by dividing by $1000$ ( $\div 1000$ ).
- Height conversions:
- Convert centimetres ( $\text{cm}$ ) to metres ( $\text{m}$ ) by dividing by $100$ ( $\div 100$ ).
- Convert millimetres ( $\text{mm}$ ) to metres ( $\text{m}$ ) by dividing by $1000$ ( $\div 1000$ ).
- Convert kilometres ( $\text{km}$ ) to metres ( $\text{m}$ ) by multiplying by $1000$ ( $\times 1000$ ).
- Energy conversions:
- Convert kilojoules ( $\text{kJ}$ ) to Joules ( $\text{J}$ ) by multiplying by $1000$ ( $\times 1000$ ).
Worked Examples for Gravitational Potential Energy:
- Example 1: Calculating Potential Energy from Mass and Height
- Given values:
  - $m = 300\,\text{g}$
  - $h = 40\,\text{m}$
  - $g = 9,8\,\text{m/s}^2$
- Unit conversion:
  - $m = \frac{300\,\text{g}}{1000} = 0,3\,\text{kg}$
- Calculation:
  - $E_p = m \times g \times h$
  - $E_p = 0,3 \times 9,8 \times 40$
  - $E_p = 117,6\,\text{J}$
- Example 2: Calculating Mass from Potential Energy and Height
- Given values:
  - $E_p = 160\,\text{J}$
  - $h = 30\,\text{m}$
  - $g = 9,8\,\text{m/s}^2$
- Calculation:
  - $E_p = m \times g \times h$
  - $160 = m \times 9,8 \times 30$
  - $160 = 294 \times m$
  - $\frac{160}{294} = \frac{294\,m}{294}$
  - $m = 0,544\,\text{kg}$
- Example 3: Calculating Height from Potential Energy and Mass
- Given values:
  - $m = 60\,\text{kg}$
  - $E_p = 0,85\,\text{kJ}$
  - $g = 9,8\,\text{m/s}^2$
- Unit conversion:
  - $E_p = 0,85\,\text{kJ} \times 1000 = 850\,\text{J}$
- Calculation:
  - $E_p = m \times g \times h$
  - $850 = 60 \times 9,8 \times h$
  - $850 = 588 \times h$
  - $\frac{850}{588} = \frac{588\,h}{588}$
  - $h = 1,446\,\text{m}$

Kinetic Energy

Definition: The energy that an object possesses due to its motion.
Fundamental Equation:
- $E_k = \frac{1}{2}\,m\,v^2$
Variables and Standard Units:
- $E_k$ : Kinetic energy, measured in Joules ( $\text{J}$ ).
- $m$ : Mass of the object in kilograms ( $\text{kg}$ ).
- $v$ : Speed or velocity of the object, measured in metres per second ( $\text{m/s}$ or $\text{m}\cdot\text{s}^{-1}$ ).
Boundary Condition for Motion:
- If an object is stationary ( $v = 0\,\text{m/s}$ , not moving), then $E_k = 0\,\text{J}$ .
Unit Conversions for Kinetic Energy Calculations:
- Velocity conversion:
- If given speed in kilometres per hour ( $\text{km/h}$ ), convert it into metres per second ( $\text{m/s}$ ) by dividing by $3,6$ ( $\div 3,6$ ).
- Mass conversion:
- Convert grams ( $\text{g}$ ) to kilograms ( $\text{kg}$ ) by dividing by $1000$ ( $\div 1000$ ).
- Energy conversion:
- Convert kilojoules ( $\text{kJ}$ ) to Joules ( $\text{J}$ ) by multiplying by $1000$ ( $\times 1000$ ).
Proportionality and Velocity Scaling:
- Kinetic energy is directly proportional to the square of the velocity ( $E_k \propto v^2$ ).
- When velocity is doubled ( $v \rightarrow 2v$ ), kinetic energy increases by a factor of four ( $4\times$ ).

Worked Examples for Kinetic Energy:
- Example 4: Velocity Doubling Derivation and Verification
- Problem formulation:
  - An object has a kinetic energy of $E$ when the velocity is $v$ . Determine the kinetic energy if the velocity doubles.
  - Options: A. $E$ , B. $2E$ , C. $4E$ , D. $\frac{1}{2}E$
  - Correct selection: C ( $4E$ )
- Algebraic proof:
  - $E_k = \frac{1}{2}\,m\,(2v)^2$
  - $E_k = \frac{1}{2}\,m\,(4v^2)$
  - $E_k = 4 \times \left(\frac{1}{2}\,m\,v^2\right) = 4E$
- Numerical demonstration:
  - Let $m = 50\,\text{kg}$ and $v = 10\,\text{m/s}$ :
  - $E_k = \frac{1}{2}\,m\,v^2 = 0,5 \times 50 \times (10)^2 = 2500\,\text{J}$
  - Doubling velocity to $v = 20\,\text{m/s}$ :
  - $E_k = 0,5 \times 50 \times (20)^2 = 0,5 \times 50 \times 400 = 10000\,\text{J}$
  - Ratio: $\frac{10000\,\text{J}}{2500\,\text{J}} = 4$
- Example 5: Calculating Kinetic Energy with Unit Conversions
- Given values:
  - $m = 800\,\text{g}$
  - $v = 9\,\text{km/h}$
- Unit conversions:
  - $m = \frac{800\,\text{g}}{1000} = 0,8\,\text{kg}$
  - $v = \frac{9\,\text{km/h}}{3,6} = 2,5\,\text{m/s}$
- Calculation:
  - $E_k = \frac{1}{2}\,m\,v^2$
  - $E_k = (0,5)(0,8)(2,5)^2$
  - $E_k = 0,4 \times 6,25$
  - $E_k = 2,5\,\text{J}$
- Example 6: Calculating Mass from Kinetic Energy and Velocity
- Given values:
  - $E_k = 9,6\,\text{kJ}$
  - $v = 20\,\text{m/s}$
- Unit conversion:
  - $E_k = 9,6\,\text{kJ} \times 1000 = 9600\,\text{J}$
- Calculation:
  - $E_k = \frac{1}{2}\,m\,v^2$
  - $9600 = \frac{1}{2}\,m\,(20)^2$
  - $9600 = \frac{1}{2}\,(400)\,m$
  - $9600 = 200\,m$
  - $\frac{9600}{200} = \frac{200\,m}{200}$
  - $m = 48\,\text{kg}$

Mechanical Energy and the Conservation Principle

Definition: Mechanical energy (also referred to as Total energy) is the sum of the gravitational potential energy and kinetic energy of an object.
Fundamental Equations:
- $E_{\text{mech}} = E_p + E_k$
- $E_{\text{mech}} = mgh + \frac{1}{2}\,m\,v^2$
Principle of Conservation of Mechanical Energy:
- Mechanical energy remains constant in an isolated system in the absence of friction or external non-conservative forces.
Strategic Rules for Energy Problem Solving:
- Work out $E_{\text{mech}}$ once per problem scenario, as it remains constant throughout the motion.
- Use the comprehensive long equation ( $E_{\text{mech}} = mgh + \frac{1}{2}\,m\,v^2$ ) whenever more than two unknown variables are present across separated energy states.
- Physical terminology convention: The word "dropped" explicitly indicates that the initial release speed of an object is zero ( $v = 0\,\text{m/s}$ ), making initial kinetic energy $E_k = 0\,\text{J}$ .

Worked mechanical energy problems for falling bodies and projected objects

Worked Examples for Mechanical Energy:
- Example 7: Dropped Object from a Building
- System configuration:
  - A ball of mass $400\,\text{g}$ is dropped from the top of a $90\,\text{m}$ building.
  - "Dropped" implies initial velocity $v = 0\,\text{m/s}$ .
  - $g = 9,8\,\text{m/s}^2$
- Unit conversion:
  - $m = \frac{400\,\text{g}}{1000} = 0,4\,\text{kg}$
- Step 7.1: Gravitational potential energy at start position:
  - $E_p = mgh$
  - $E_p = 0,4 \times 9,8 \times 90$
  - $E_p = 352,8\,\text{J}$
- Step 7.2: Kinetic energy at start position:
  - $E_k = 0\,\text{J}$
  - Explanation: The ball is not moving at the release point.
- Step 7.3: Total mechanical energy at start position:
  - $E_{\text{mech}} = E_p + E_k$
  - $E_{\text{mech}} = 352,8 + 0$
  - $E_{\text{mech}} = 352,8\,\text{J}$
- Example 8: Object Possessing Both Potential and Kinetic Energy
- System configuration:
  - A ball of mass $600\,\text{g}$ is at a height of $30\,\text{m}$ above the ground.
  - Speed of the ball: $v = 3\,\text{m/s}$ .
  - $g = 9,8\,\text{m/s}^2$
- Unit conversion:
  - $m = \frac{600\,\text{g}}{1000} = 0,6\,\text{kg}$
- Step 8.1: Potential energy at this height:
  - $E_p = mgh$
  - $E_p = 0,6 \times 9,8 \times 30$
  - $E_p = 176,4\,\text{J}$
- Step 8.2: Kinetic energy at this height:
  - $E_k = \frac{1}{2}\,m\,v^2$
  - $E_k = 0,5 \times 0,6 \times (3)^2$
  - $E_k = 0,3 \times 9$
  - $E_k = 2,7\,\text{J}$
- Step 8.3: Mechanical energy at this height:
  - $E_{\text{mech}} = E_p + E_k$
  - $E_{\text{mech}} = 176,4 + 2,7$
  - $E_{\text{mech}} = 179,1\,\text{J}$
- Example 9: Dropped Cellphone and Conservation of Energy
- System configuration:
  - A man drops a cellphone of mass $200\,\text{g}$ from a window $40\,\text{m}$ above the ground.
  - $g = 9,8\,\text{m/s}^2$
- Unit conversion:
  - $m = \frac{200\,\text{g}}{1000} = 0,2\,\text{kg}$
- Step 9.1: Kinetic energy of the phone at the window:
  - $E_k = \frac{1}{2}\,m\,v^2 = 0,5 \times 0,2 \times (0)^2 = 0\,\text{J}$
  - Explanation: The phone is not moving at the instant of release.
- Step 9.2: Potential energy of the phone at the window:
  - $E_p = mgh$
  - $E_p = 0,2 \times 9,8 \times 40$
  - $E_p = 78,4\,\text{J}$
- Step 9.3: Mechanical energy of the phone at the window:
  - $E_{\text{mech}} = E_p + E_k$
  - $E_{\text{mech}} = 78,4 + 0$
  - $E_{\text{mech}} = 78,4\,\text{J}$
- Step 9.4: Mechanical energy after falling $15\,\text{m}$ downwards:
  - $E_{\text{mech}} = 78,4\,\text{J}$
  - Explanation: Mechanical energy remains constant throughout free fall in the absence of friction.

Practice Problems and Energy Worksheet

Problem 1:
- The speed of a $3\,\text{kg}$ object is $18\,\text{km/h}$ .
- Objective: Calculate its kinetic energy.
- Necessary conversion: Convert $18\,\text{km/h}$ into $\text{m/s}$ by dividing by $3,6$ : $\frac{18}{3,6} = 5\,\text{m/s}$ .
Problem 2:
- The speed of a $1400\,\text{g}$ ball is $4\,\text{m/s}$ .
- Objective: Calculate its kinetic energy.
- Necessary conversion: Convert $1400\,\text{g}$ to kilograms: $\frac{1400}{1000} = 1,4\,\text{kg}$ .
Problem 3:
- A $2\,\text{kg}$ object is thrown downwards in the air at $3\,\text{m/s}$ from a $60\,\text{m}$ ledge.
- Sub-problems:
- 3.1 Calculate the potential energy of the object at this position.
- 3.2 Calculate the kinetic energy of the object at this position.
- 3.3 Calculate the mechanical energy of the object at this position.
- 3.4 Explain why the speed of the object increases as it travels downwards.
- 3.5 The object descends to reach a new height of $24\,\text{m}$ :
  - 3.5.1 Will its potential energy increase, decrease, or stay the same?
  - 3.5.2 Will its kinetic energy increase, decrease, or stay the same?
  - 3.5.3 Will its mechanical energy increase, decrease, or stay the same?
Problem 4:
- A stone has a mass of $80\,\text{g}$ at a height of $260\,\text{cm}$ above the roof of a car.
- Objective: Calculate its potential energy at this position.
- Necessary conversions:
- Mass: $\frac{80\,\text{g}}{1000} = 0,08\,\text{kg}$
- Height: $\frac{260\,\text{cm}}{100} = 2,6\,\text{m}$
Problem 5:
- The potential energy of a stationary $4\,\text{kg}$ object is $0,8\,\text{kJ}$ at a height $h$ .
- Sub-problems:
- 5.1 Calculate the height $h$ .
- 5.2 Determine the weight ( $W = m \times g$ ) of this object.
- 5.3 Calculate the mechanical energy of this object.
Problem 6:
- A metal ball has a weight of $588\,\text{N}$ at a height of $400\,\text{cm}$ above the ground and is moving at a velocity of $2\,\text{m/s}$ .
- Sub-problems:
- 6.1 Calculate the mass of the ball (using $W = mg$ , where $g = 9,8\,\text{m/s}^2$ ).
- 6.2 Calculate the potential energy of the ball at this height.
- 6.3 Calculate the mechanical energy of the ball at this height.

Gravitational Potential Energy

Definition: Energy an object possesses due to its position above a reference point (usually the ground).
Formula:
- $E_p = mgh$
Variables and Standard Units:
- $E_p$ : Potential energy in Joules ( $\text{J}$
- $m$ : Mass in kilograms ( $\text{kg}$
- $g$ : Gravitational acceleration ( $9.8\,\text{m/s}^2$ on Earth)
- $h$ : Height above the reference point in metres ( $\text{m}$
Key Rules and Proportionalities:
- If $h = 0\,\text{m}$ , then $E_p = 0\,\text{J}$ .
- $E_p$ is directly proportional to both mass ( $m$ ) and height ( $h$ ).
Rearranged Formulas:
- $m = \frac{E_p}{gh}$
- $h = \frac{E_p}{mg}$
- $g = \frac{E_p}{mh}$
Unit Conversions:
- Mass: Grams ( $\text{g}$ ) to kilograms ( $\text{kg}$ ) $\rightarrow \div 1000$
- Height:
- Centimetres ( $\text{cm}$ ) to metres ( $\text{m}$ ) $\rightarrow \div 100$
- Millimetres ( $\text{mm}$ ) to metres ( $\text{m}$ ) $\rightarrow \div 1000$
- Kilometres ( $\text{km}$ ) to metres ( $\text{m}$ ) $\rightarrow \times 1000$
- Energy: Kilojoules ( $\text{kJ}$ ) to Joules ( $\text{J}$ ) $\rightarrow \times 1000$

Kinetic Energy

Definition: Energy an object possesses due to its motion.
Formula:
- $E_k = \frac{1}{2}\,m\,v^2$
Variables and Standard Units:
- $E_k$ : Kinetic energy in Joules ( $\text{J}$
- $m$ : Mass in kilograms ( $\text{kg}$
- $v$ : Speed or velocity in metres per second ( $\text{m/s}$
Key Rules and Proportionalities:
- If stationary ( $v = 0\,\text{m/s}$ ), then $E_k = 0\,\text{J}$ .
- $Ek \propto v^2$ : Doubling velocity ( $2v$ ) quadruples kinetic energy ( $4 \times Ek$ ).
Unit Conversions:
- Speed: $\text{km/h}$ to $\text{m/s}$ $\rightarrow \div 3.6$
- Mass: $\text{g}$ to $\text{kg}$ $\rightarrow \div 1000$
- Energy: $\text{kJ}$ to $\text{J}$ $\rightarrow \times 1000$

Mechanical Energy and Conservation

Definition: The total energy of an object, equal to the sum of its potential and kinetic energy.
Formula:
- $E{\text{mech}} = Ep + E_k$
- $E_{\text{mech}} = mgh + \frac{1}{2}\,m\,v^2$
Conservation Principle: In an isolated system (no friction or external forces), total mechanical energy remains constant throughout the motion.
Problem-Solving Guidelines:
- Calculate $E_{\text{mech}}$ once per scenario, as it does not change.
- The word "dropped" means initial speed is zero ( $v = 0\,\text{m/s}$ ), so initial $E_k = 0\,\text{J}$ .
- As an object falls, $Ep$ decreases and converts into $Ek$ , causing speed to increase while $E_{\text{mech}}$ stays constant.

Practice Problem Prompts

Calculate $E_k$ for a $3\,\text{kg}$ object moving at $18\,\text{km/h}$ ( $\text{convert: } 18 \div 3.6 = 5\,\text{m/s}$ .
Calculate $E_k$ for a $1400\,\text{g}$ ball moving at $4\,\text{m/s}$ ( $\text{convert: } 1400 \div 1000 = 1.4\,\text{kg}$ .
A $2\,\text{kg}$ object is thrown downwards at $3\,\text{m/s}$ from a height of $60\,\text{m}$ :
- Find $Ep$ , $Ek$ , and $E_{\text{mech}}$ .
- Explain changes in $Ep$ , $Ek$ , and $E_{\text{mech}}$ as it falls to $24\,\text{m}$ .
Calculate $E_p$ of an $80\,\text{g}$ stone at a height of $260\,\text{cm}$ .
A stationary $4\,\text{kg}$ object has $E_p = 0.8\,\text{kJ}$ :
- Calculate height $h$ , weight ( $W = mg$ ), and $E_{\text{mech}}$ .
A ball with a weight of $588\,\text{N}$ is $400\,\text{cm}$ high and moving at $2\,\text{m/s}$ :
- Calculate mass, $Ep$ , and $E{\text{mech}}$ .