### Question 20 of 25The potential energy of an object attached to a spring is 2.70 J at a location where the kinetic energy is 1.20 J. If the amplitude \( A \) of the simple harmonic motion is 19.0 cm, calculate the spring constant \( k \) and the magnitude of the largest force \( F_{\text{spring, max}} \) that the object experiences.---\[ k = \quad \text{N} \hspace{-1mm}/\hspace{-1mm} \text{m} \]\[ F_{\text{spring, max}} = \quad \text{N} \]---In this problem, we need to determine two quantities related to a spring undergoing simple harmonic motion: the spring constant \( k \) and the maximum spring force \( F_{\text{spring, max}} \).#### Given Data:- Potential Energy \( U = 2.70 \) J- Kinetic Energy \( K = 1.20 \) J- Amplitude \( A = 19.0 \) cm (which is 0.190 m when converted to meters)#### Relevant Equations:1. The total mechanical energy in the system is the sum of the potential and kinetic energies: \[ E_{\text{total}} = U + K \]2. The total mechanical energy can also be expressed in terms of the amplitude \( A \) and the spring constant \( k \): \[ E_{\text{total}} = \frac{1}{2} k A^2 \]3. The maximum force exerted by a spring \( F_{\text{spring, max}} \) can be found using Hooke's Law: \[ F_{\text{spring, max}} = k A \]#### Steps to Solve:1. Calculate the total mechanical energy: \[ E_{\text{total}} = 2.70 \, \text{J} + 1.20 \, \text{J} = 3.90 \, \text{J} \]2. Use the total mechanical energy equation to solve for the spring constant \( k \): \[ 3.90 \, \text{J} = \frac{1}{2} k (0.190 \, \text{m})

Name: ### Question 20 of 25The potential energy of an object attached to a
Uploaded: 2024-03-20T06:58:04.000Z
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Description: ### Question 20 of 25The potential energy of an object attached to a spring is 2.70 J at a location where the kinetic energy is 1.20 J. If the amplitude \( A \) of the simple harmonic motion is 19.0 cm, calculate the spring constant \( k \) and the