Sure, let's break this down step-by-step. You are using a spring-mass system, changing gravity (Jupiter/Earth), and using a 100g mass. Let’s go step by step: --- ### **Step 1: Calculate the Spring Stretch on Jupiter** **Given:** - Mass, \( m = 100\,g = 0.1\,kg \) - Gravity on Jupiter, \( g_{Jupiter} = 24.79\,m/s^2 \) - Spring stretches to a new equilibrium where \( F_{spring} = F_{gravity} \) - \( F_{gravity} = m \times g \) - \( F_{spring} = kx \), where \( x \) is the stretch from natural length, \( k \) is the spring constant **Calculation:** \[ F_{gravity} = m \cdot g = 0.1 \cdot 24.79 = 2.479\,N \] \[ k = \text{(You select, say 7 N/m or 9 N/m; let's use 7 N/m first)} \] \[ k x = 2.479 \implies x = \frac{2.479}{7} = 0.354\,m = 35.4\,cm \] **Explanation:** On Jupiter, the force due to gravity is much higher, so the spring stretches more. For a spring constant of 7 N/m, the stretch is **35.4 cm**. --- ### **Step 2: What is the Spring Constant?** **Given:** - You already set the spring constant in the simulation (either 7 N/m or 9 N/m according to your selection). **Explanation:** The spring constant \( k \) is a property of the spring and does **not** depend on gravity. It is set by you in the simulation. **If you set 7 on the scale, the spring constant is 7 N/m. If you set 9, it is 9 N/m.** --- ### **Step 3: Change Gravity to Earth, Find the Stretch** **Given:** - Gravity on Earth, \( g_{Earth} = 9.8\,m/s^2 \) \[ F_{gravity} = 0.1 \cdot 9.8 = 0.98\,N \] \[ k x = 0.98 \implies x = \frac{0.98}{7} = 0.14\,m = 14.0\,cm \] **Explanation:** On Earth, with lower gravity, the spring stretches less. For \( k = 7\,N/m \), the stretch is **14.0 cm**. --- ### **Step 4: What is the Spring Constant on Earth?** **Explanation:** The spring constant is **the same** as before, since you did not change the spring. If it was 7 N/m on Jupiter, it stays 7 N/m on Earth. --- ### **Step 5: Compare the Spring Constants (Question 5)** **Explanation:** The spring constant does **not** depend on gravity or location. It is a property of the spring itself. **Correct answer is:** > **B. They should be the same as the spring did not change and the spring constant doesn't depend on gravity.** --- ## **Final Answers:** **1. How much was the spring stretched from original length (on Jupiter, k = 7 N/m)?** - **35.4 cm** **2. What is the spring constant?** - **7 N/m** (or whatever value you set) **3. How much was the spring stretched (on Earth, k = 7 N/m)?** - **14.0 cm** **4. What is the spring constant?** - **7 N/m** (same as before) **5. How is your spring constant in question 1 compared to question 3?** - **B. They should be the same as the spring did not change and the spring constant doesn't depend on gravity.** --- If you used a spring constant of 9 N/m, plug 9 into the calculations above for the stretches.

The fundamental concept underlying these questions is that the spring constant (k) is an intrinsic property of the spring itself and remains unchanged regardless of external factors like gravity. When a mass is attached to a spring and displaced, the spring stretches until the restoring force balances the weight of the mass, reaching an equilibrium position. This equilibrium stretch depends on the applied force (which is the weight of the mass, \( m \times g \)) and the spring constant, following Hooke's Law (\( F = kx \)). Since gravity influences the weight of the mass and thus the force applied to the spring, the amount of stretch will vary with gravity if the same spring and mass are used. However, the spring constant remains constant because it is determined by the spring's material and geometry, not by external forces. Therefore, when gravity changes (e.g., from Jupiter to Earth), the equilibrium stretch changes proportionally to the gravitational acceleration, but the spring constant stays the same. This demonstrates that the spring's stiffness is independent of the gravitational environment, emphasizing the distinction between a spring's intrinsic properties and the forces acting upon it.