Summary of Work, Energy, and Power

## Introduction An object's mechanical energy is the sum of its **kinetic energy** and its **gravitational potential energy**. In many frictionless situations, energy transforms between these two forms while keeping the total mechanical energy constant. > **Definition:** Kinetic energy is the energy associated with an object's motion. Gravitational potential energy is the energy stored due to an object's vertical position in a gravitational field. ## Fundamental Concepts ### Kinetic Energy (KE) > **Definition:** The kinetic energy of an object with mass $m$ moving at speed $v$ is given by $$\text{KE} = \frac{1}{2} m v^{2}$$ - Unit: joules (J). - It is a scalar quantity: only speed matters, not direction. Example: An object with a mass of $2\,\mathrm{kg}$ moving at a speed of $3\,\mathrm{m/s}$ has $$\text{KE} = \frac{1}{2} \times 2\,\mathrm{kg} \times (3\,\mathrm{m/s})^{2} = 9\,\mathrm{J}$$ Effect of doubling the speed: If $v$ is doubled, $\text{KE}$ is multiplied by $4$ because it depends on $v^{2}$. ### Gravitational Potential Energy (PE) > **Definition:** The gravitational potential energy of an object with mass $m$ located at a height $h$ above a reference point is $$\text{PE} = m g h$$ - $g$ is the acceleration due to gravity; here we will use $g = 10\,\mathrm{N/kg}$ when specified. - Unit: joules (J). Example: An object with a mass of $2\,\mathrm{kg}$ at a height of $3\,\mathrm{m}$ with $g=10\,\mathrm{N/kg}$ has $$\text{PE} = 2\,\mathrm{kg} \times 10\,\mathrm{N/kg} \times 3\,\mathrm{m} = 60\,\mathrm{J}$$ ### Conservation of Mechanical Energy In the absence of dissipative forces (friction, air resistance), the loss in PE is equal to the gain in KE: $$\Delta \text{PE} + \Delta \text{KE} = 0$$ or equivalently $$\text{PE}_{i} + \text{KE}_{i} = \text{PE}_{f} + \text{KE}_{f}$$ This allows us to solve problems without directly using the equations of motion. ## Step-by-Step Breakdown and Practical Examples ### 1) Calculating Potential Energy (PE) for a $6\,\mathrm{kg}$ Mass - a) At $4\,\mathrm{m}$: $$\text{PE} = m g h = 6\,\mathrm{kg} \times 10\,\mathrm{N/kg} \times 4\,\mathrm{m} = 240\,\mathrm{J}$$ - b) At $6\,\mathrm{m}$: $$\text{PE} = 6\,\mathrm{kg} \times 10\,\mathrm{N/kg} \times 6\,\mathrm{m} = 360\,\mathrm{J}$$ ### 2) Kinetic Energy (KE) and the Effect of Doubling Velocity - For a mass of $6\,\mathrm{kg}$ and a velocity of $5\,\mathrm{m/s}$: $$\text{KE} = \frac{1}{2} \times 6\,\mathrm{kg} \times (5\,\mathrm{m/s})^{2} = \frac{1}{2} \times 6 \times 25 = 75\,\mathrm{J}$$ - If the velocity is doubled ($v = 10\,\mathrm{m/s}$): $$\text{KE} = \frac{1}{2} \times 6\,\mathrm{kg} \times (10\,\mathrm{m/s})^{2} = \frac{1}{2} \times 6 \times 100 = 300\,\mathrm{J}$$ (Notice that the KE is multiplied by 4 when $v$ is doubled.) ### 3) Example with a $0.5\,\mathrm{kg}$ Ball and $100\,\mathrm{J}$ of KE To find velocity, use $\text{KE} = \tfrac{1}{2} m v^{2}$ $$100\,\mathrm{J} = \frac{1}{2} \times 0.5\,\mathrm{kg} \times v^{2}$$ Solve for $v$: $$100 = 0.25 \; v^{2}$$ $$v^{2} = 400$$ $$v = 20\,\mathrm{m/s}$$ ### 4) A $0.5\,\mathrm{kg}$ Ball Dropped from a Cliff - Upon impact with a speed of $10\,\mathrm{m/s}$, its final KE is: $$\text{KE} = \frac{1}{2} \times 0.5\,\mathrm{kg} \times (10\,\mathrm{m/s})^{2} = 25\,\mathrm{J}$$ - If there is no friction, this final KE comes from the initial PE; therefore, the PE before being dropped was $25\,\mathrm{J}$. - Initial height (using $\text{PE} = m g h$): $$25\,\mathrm{J} = 0.5\,\mathrm{kg} \times 10\,\mathrm{N/kg} \times h$$ $$h = \frac{25}{5} = 5\,\mathrm{m}$$ ### 5) A $1\,\mathrm{kg}$ Stone Impacting at $10\,\mathrm{m/s}$ - Final KE: $$\text{KE} = \frac{1}{2} \times 1\,\mathrm{kg} \times (10\,\mathrm{m/s})^{2} = 50\,\mathrm{J}$$ - Initial PE = $50\,\mathrm{J}$. - Height of fall: $$50 = 1 \times 10 \times h$$ $$h = 5\,\mathrm{m}$$ ### Worked Example from the Provided Materials A $4\,\mathrm{kg}$ stone at the top of a $5\,\mathrm{m}$ height: PE $=200\,\mathrm{J}$. If all the PE is convert