Certainly! Let’s tackle **Question 1(a)** step-by-step, as an example, and then I’ll briefly outline steps for the other parts. If you’d like detailed step-by-steps for all, let me know! --- ## 1. Roller Coaster and Spring Problem ### (a) **Find the spring stiffness \( k \) so that the car barely makes it around the loop without falling.** #### **Step 1: Find the minimum speed at the top of the loop** - At the top of the loop, for the car to just make it, the **normal force** is zero. Only gravity provides the required centripetal force: \[ mg = \frac{mv^2}{r} \implies v_{\text{min}} = \sqrt{gr} \] where \( r = 10\,\text{m} \) (radius of loop). **Explanation:** At the top of the loop, the only force providing the centripetal acceleration is gravity. Setting the normal force to zero gives the minimum speed needed to stay on the track. #### **Step 2: Apply energy conservation from the start to the top of the loop** - Initial energy (compressed spring): \[ E_{\text{initial}} = \frac{1}{2} k x^2 \] - Final energy at the top of the loop (height = \(2r\)): \[ E_{\text{final}} = mgh + \frac{1}{2}mv^2 \] where \( h = 2r \), \( v = \sqrt{gr} \). - Set initial and final energy equal: \[ \frac{1}{2}k x^2 = mg(2r) + \frac{1}{2}m(gr) \] Substitute \( m = 500\,\text{kg} \), \( r = 10\,\text{m} \), \( g = 9.8\,\text{m/s}^2 \), \( x = 6\,\text{m} \): \[ \frac{1}{2}k (6)^2 = 500 \times 9.8 \times (20) + \frac{1}{2} \times 500 \times (9.8 \times 10) \] \[ 18k = 98,000 + 24,500 \] \[ 18k = 122,500 \] \[ k = \frac{122,500}{18} \approx 6,806\,\text{N/m} \] **Explanation:** The spring must provide enough energy to lift the car to the top of the loop and give it the minimum speed to stay on the track. --- ### Final Answer for 1(a): \[ \boxed{k \approx 6,806\,\text{N/m}} \] --- If you want the same **2-step breakdown** for parts 1(b), 1(c), and 1(d), just let me know which part to do next!

Certainly! Let's solve the entire problem step-by-step. --- ## 1. Roller Coaster and Spring Problem ### (a) **Find the spring stiffness \( k \) so that the car barely makes it around the loop without falling.** #### **Step 1: Find the minimum speed at the top of the loop** At the top of the loop, for the car to just make it, the normal force is zero. Gravity alone provides the centripetal force: \[ mg = \frac{mv^2}{r} \implies v_{\text{min}} = \sqrt{gr} \] where \( r = 10\,\text{m} \). - **Explanation:** At the top of the loop, the only force providing the centripetal acceleration is gravity. Setting the normal force to zero gives the minimum speed needed to stay on the track. #### **Step 2: Apply energy conservation from the start to the top of the loop** Initial energy (compressed spring): \[ E_{\text{initial}} = \frac{1}{2} k x^2 \] Final energy at the top of the loop (height = \( 2r \)): \[ E_{\text{final}} = mgh + \frac{1}{2}mv^2 \] where \( h = 2r \), \( v = \sqrt{gr} \). Set initial and final energy equal: \[ \frac{1}{2}k x^2 = mg(2r) + \frac{1}{2}m(gr) \] Substituting \( m = 500\,\text{kg} \), \( r = 10\,\text{m} \), \( g = 9.8\,\text{m/s}^2 \), \( x = 6\,\text{m} \): \[ \frac{1}{2}k (6)^2 = 500 \times 9.8 \times 20 + \frac{1}{2} \times 500 \times (9.8 \times 10) \] \[ 18k = 98,000 + 24,500 \] \[ 18k = 122,500 \implies k = \frac{122,500}{18} \approx 6,806\,\text{N/m} \] - **Explanation:** The spring must provide enough energy to lift the car to the top of the loop and give it the minimum speed to stay on the track. --- ### Final Answer for 1(a): \[ \boxed{k \approx 6,806\,\text{N/m}} \] --- ### (b) **Calculate the force exerted by the spring when the car is at the bottom of the loop.** #### **Step 1: Find the total energy at the bottom of the loop** At the bottom of the loop, the potential energy is zero, and the kinetic energy can be calculated using: \[ E = \frac{1}{2} k x^2 \implies KE = \frac{1}{2} mv^2 \] #### **Step 2: Set the energies equal** \[ \frac{1}{2} k x^2 = \frac{1}{2} mv^2 \] Using \( k = 6806\,\text{N/m} \) and the found \( v \): \[ F = kx \] Substituting \( x = 6\,\text{m} \): \[ F = 6806 \times 6 = 40,836\,\text{N} \] - **Explanation:** The force exerted by the spring is calculated using Hooke's law, where the spring's compression provides the force when the car is at the bottom of the loop. --- ### Final Answer for 1(b): \[ \boxed{F \approx 40,836\,\text{N}} \] --- ### (c) **Determine the maximum speed of the car at the bottom of the loop.** #### **Step 1: Use conservation of energy** The total mechanical energy at the top equals the total mechanical energy at the bottom: \[ E_{\text{top}} = E_{\text{bottom}} \implies mgh + \frac{1}{2}mv_{\text{top}}^2 = \frac{1}{2}mv_{\text{bottom}}^2 \] #### **Step 2: Solve for \( v_{\text{bottom}} \)** Using \( h = 2r \): \[ m \cdot g \cdot 20 = \frac{1}{2} m v_{\text{bottom}}^2 \] \[ v_{\text{bottom}} = \sqrt{40g} = \sqrt{40 \times 9.8} \approx 19.8\,\text{m/s} \] - **Explanation:** The maximum speed at the bottom is derived from energy conservation, where potential energy at the top converts to kinetic energy at the bottom. --- ### Final Answer for 1(c): \[ \boxed{v_{\text{bottom}} \approx 19.8\,\text{m/s}} \] --- ### (d) **Calculate the acceleration of the car at the bottom of the loop.** #### **Step 1: Centripetal acceleration** The centripetal acceleration \( a_c \) is given by: \[ a_c = \frac{v^2}{r} \] #### **Step 2: Substitute the values** Substituting \( v \) and \( r \): \[ a_c = \frac{(19.8)^2}{10} \approx 39.2\,\text{m/s}^2 \] - **Explanation:** The acceleration at the bottom of the loop is centripetal, caused by the change in direction of the car as it moves in a circular path. --- ### Final Answer for 1(d): \[ \boxed{a_c \approx 39.2\,\text{m/s}^2} \] --- This completes the solution for all parts of the problem! If you have any more questions or need further clarification, feel free to ask!