Let's break this down step by step! --- ## **Definitions** ### **1. What is a Newton?** A **Newton (N)** is the SI unit of force. **1 Newton** is the amount of force required to accelerate a **1 kilogram mass by 1 meter per second squared** (1 N = 1 kg·m/s²). --- ### **2. What is a Joule?** A **Joule (J)** is the SI unit of work or energy. **1 Joule** is the work done when a force of **1 Newton moves an object 1 meter** (1 J = 1 N·m). --- ### **3. What is a Watt?** A **Watt (W)** is the SI unit of power. **1 Watt** is equal to **1 Joule of work done per second** (1 W = 1 J/s). --- ## **Question #1: Force Needed (Newtons)** **Given:** - Car #1 (light truck): 2,000 kg, acceleration = 5 m/s² - Car #2 (subcompact): 1,000 kg, acceleration = 5 m/s² **Formula:** Force (F) = mass (m) × acceleration (a) ### **Car #1:** \( F_1 = 2000\,kg \times 5\,m/s^2 = 10,000\,N \) ### **Car #2:** \( F_2 = 1000\,kg \times 5\,m/s^2 = 5,000\,N \) **Which car uses less force?** **Car #2** uses less force because it has less mass; for the same acceleration, less mass needs less force. --- ## **Question #2: Work Done (Joules)** **Formula:** Work (W) = Force (F) × distance (d) **Given:** distance = 100 m ### **Car #1:** \( W_1 = 10,000\,N \times 100\,m = 1,000,000\,J \) ### **Car #2:** \( W_2 = 5,000\,N \times 100\,m = 500,000\,J \) --- ## **Question #3: Power Used (Watts)** **Formula:** Power (P) = Work (W) / time (t) **Given:** time = 10 s ### **Car #1:** \( P_1 = \frac{1,000,000\,J}{10\,s} = 100,000\,W \) ### **Car #2:** \( P_2 = \frac{500,000\,J}{10\,s} = 50,000\,W \) --- ## **Question #4: Daily Mitigation Cost** **Convert Watts to kilowatts (kW):** - \( P_1 = 100,000\,W = 100\,kW \) - \( P_2 = 50,000\,W = 50\,kW \) **Cost:** $0.10 per kW per day ### **Car #1:** \( 100\,kW \times \$0.10 = \$10.00 \) per day ### **Car #2:** \( 50\,kW \times \$0.10 = \$5.00 \) per day --- ## **Question #5: Annual Mitigation Cost (365 days)** ### **Car #1:** \( \$10.00 \times 365 = \$3,650 \) per year ### **Car #2:** \( \$5.00 \times 365 = \$1,825 \) per year --- ## **Question #6: Difference in Environmental Cost** \( \$3,650 - \$1,825 = \$1,825 \) **Car #1 costs $1,825 more per year than Car #2 in environmental mitigation.** --- ## **Summary Table** | | Car #1 (Truck) | Car #2 (Subcompact) | |----------------|------------------|---------------------| | Force (N) | 10,000 | 5,000 | | Work (J) | 1,000,000 | 500,000 | | Power (W) | 100,000 | 50,000 | | Power (kW) | 100 | 50 | | Daily Cost ($) | 10.00 | 5.00 | | Annual Cost ($)| 3,650 | 1,825 | | Difference ($) | \-\- | 1,825 | --- **Let me know if you need any explanations or further clarifications!**

Let's analyze and solve each part of the problem step by step. --- ### **Given Data:** - Spring constant, \(k = 45\,N/m\) - Initial stretch, \(x_{initial} = 6.0\,cm = 0.06\,m\) - Final stretch when the spring returns to equilibrium: \(x = 0\,m\) - Mass of the ball, \(m = 500\,g = 0.5\,kg\) --- ### **a. Change in potential energy of the spring from initial stretch to equilibrium** **Potential energy stored in a stretched/compressed spring:** \[ PE_{spring} = \frac{1}{2} k x^2 \] - Initial potential energy at \(x_{initial} = 0.06\,m\): \[ PE_{initial} = \frac{1}{2} \times 45\,N/m \times (0.06\,m)^2 \] \[ PE_{initial} = 0.5 \times 45 \times 0.0036 = 0.5 \times 0.162 = 0.081\,J \] - Potential energy at equilibrium position (\(x=0\)): \[ PE_{final} = 0\,J \] **Change in potential energy:** \[ \Delta PE = PE_{final} - PE_{initial} = 0 - 0.081\,J = -0.081\,J \] **Answer:** \[ \boxed{ \Delta PE = -0.081\,J } \] (It's negative because the spring loses potential energy as it returns to equilibrium.) --- ### **b. Speed of the ball when the spring returns to its equilibrium length** **Principle:** The initial potential energy in the spring converts into kinetic energy at equilibrium (assuming no friction). \[ PE_{spring} \text{ at } x=0.06\,m \rightarrow KE \text{ at } x=0 \] **Using conservation of energy:** \[ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \] Solve for \(v\): \[ v = \sqrt{\frac{k x^2}{m}} \] Plugging in the values: \[ v = \sqrt{\frac{45\,N/m \times (0.06\,m)^2}{0.5\,kg}} = \sqrt{\frac{45 \times 0.0036}{0.5}} \] Calculate numerator: \[ 45 \times 0.0036 = 0.162 \] Calculate \(v\): \[ v = \sqrt{\frac{0.162}{0.5}} = \sqrt{0.324} \approx 0.569\,m/s \] **Answer:** \[ \boxed{ v \approx 0.569\,m/s } \] --- ### **c. Speed of the ball when the spring is stretched by 3.0 cm as it returns from 6.0 cm** **Scenario:** The spring is initially stretched by 6.0 cm (\(x_{initial} = 0.06\,m\)) and is allowed to pass through the equilibrium point and reach a stretch of 3.0 cm (\(x_{final} = 0.03\,m\)). **Conservation of energy:** Initial potential energy at \(x_{initial} = 0.06\,m\): \[ PE_{initial} = \frac{1}{2} \times 45 \times (0.06)^2 = 0.081\,J \] Potential energy at \(x_{final} = 0.03\,m\): \[ PE_{final} = \frac{1}{2} \times 45 \times (0.03)^2 = 0.5 \times 45 \times 0.0009 = 0.5 \times 0.0405 = 0.02025\,J \] **Kinetic energy at \(x=0.03\,m\):** \[ KE = PE_{initial} - PE_{final} = 0.081 - 0.02025 = 0.06075\,J \] Calculate the speed: \[ KE = \frac{1}{2} m v^2 \Rightarrow v = \sqrt{\frac{2 KE}{m}} \] \[ v = \sqrt{\frac{2 \times 0.06075}{0.5}} = \sqrt{\frac{0.1215}{0.5}} = \sqrt{0.243} \approx 0.493\,m/s \] **Answer:** \[ \boxed{ v \approx 0.493\,m/s } \] --- ### **Summary:** | Part | Result | |--------|---------| | a | \(-0.081\,J\) (energy decreases) | | b | \(\approx 0.569\,m/s\) | | c | \(\approx 0.493\,m/s\) | Let me know if you'd like further clarification!