Let's carefully analyze the problem and find the amplitude of the motion step-by-step. --- ### **Given Data:** - Mass of the block, \( m = 0.300 \, \text{kg} \) - Position of the block, \( x = 40.240 \, \text{m} \) - Acceleration at this position, \( a_p = -14.0 \, \text{m/s}^2 \) - Velocity at this position, \( v_r = 400 \, \text{m/s} \) --- ### **Objective:** Find the **amplitude** \( A \) of the oscillation. --- ### **Step 1: Understanding the physical model** Since the surface is frictionless and the spring is ideal (Hooke's law applies), the motion is simple harmonic motion (SHM). The key equations for SHM are: \[ a = -\frac{k}{m} x \] and the total energy: \[ E = \frac{1}{2} m v^2 + \frac{1}{2} k x^2 \] where: - \( k \) is the spring constant, - \( x \) is the displacement from equilibrium position, - \( v \) is the velocity, - \( a \) is the acceleration. --- ### **Step 2: Find the spring constant \(k\)** From the given acceleration: \[ a = -\frac{k}{m} x \] At the position \( x = 40.240\, \text{m} \): \[ a_p = -14.0\, \text{m/s}^2 \] Plugging in: \[ -14.0 = -\frac{k}{0.300} \times 40.240 \] Solve for \(k\): \[ 14.0 = \frac{k}{0.300} \times 40.240 \] \[ k = \frac{14.0 \times 0.300}{40.240} \] Calculate numerator: \[ 14.0 \times 0.300 = 4.2 \] Now: \[ k = \frac{4.2}{40.240} \approx 0.1044\, \text{N/m} \] --- ### **Step 3: Find the total energy \(E\) at this point** Using the energy conservation: \[ E = \frac{1}{2} m v^2 + \frac{1}{2} k x^2 \] At the given point: \[ v = 400\, \text{m/s} \] Compute kinetic energy: \[ KE = \frac{1}{2} \times 0.300 \times (400)^2 = 0.15 \times 160,000 = 24,000\, \text{J} \] Calculate potential energy: \[ PE = \frac{1}{2} \times 0.1044 \times (40.240)^2 \] Calculate \( (40.240)^2 \): \[ 40.240^2 \approx 1619.68 \] Now: \[ PE \approx 0.0522 \times 1619.68 \approx 84.6\, \text{J} \] Total energy: \[ E \approx 24,000 + 84.6 \approx 24,084.6\, \text{J} \] --- ### **Step 4: Express total energy at an arbitrary position** In SHM, total energy is constant and can be written as: \[ E = \frac{1}{2} k A^2 \] where \(A\) is the amplitude (the maximum displacement). At the maximum displacement (amplitude), velocity is zero, and potential energy is maximum: \[ E = \frac{1}{2} k A^2 \] At any position \(x\), the energy can also be written as: \[ E = \frac{1}{2} m v^2 + \frac{1}{2} k x^2 \] --- ### **Step 5: Find the amplitude \(A\)** At the maximum displacement \(A\), the kinetic energy is zero, so: \[ E = \frac{1}{2} k A^2 \] From earlier, total energy \(E \approx 24,084.6\, \text{J}\), so: \[ A = \sqrt{\frac{2E}{k}} = \sqrt{\frac{2 \times 24,084.6}{0.1044}} \] Calculate numerator: \[ 2 \times 24,084.6 \approx 48,169.2 \] Now: \[ A = \sqrt{\frac{48,169.2}{0.1044}} \approx \sqrt{460,916} \] Calculate the square root: \[ A \approx 678.75\, \text{m} \] --- ### **### **Final Answer:** \[ \boxed{ \text{The amplitude of the motion is approximately } \boxed{679\, \text{meters}} } \] --- ### **Note:** This is an extremely large amplitude, which is physically unusual for typical spring-mass systems, but based on the given data, mathematically, this is the result. --- **Summary:** 1. Calculated the spring constant \(k \approx 0.1044\, \text{N/m}\). 2. Used the energy conservation to find total energy \(E \approx 24,084.6\, \text{J}\). 3. Derived the amplitude \(A\) from the energy expression \(A = \sqrt{\frac{2E}{k}}\). 4. Final result: **approximately 679 meters**.