https://file.aiquickdraw.com/mathbot/file/17302588040141xnwm2sq.jpg

Let's solve this problem by using the principles of energy conservation in simple harmonic motion.

### Problem Summary:
- Mass \( m = 0.400 \, \text{kg} \)
- Spring constant \( k = 280 \, \text{N/m} \)
- Position \( x = 0.200 \, \text{m} \)
- Velocity at \( x = 0.200 \, \text{m} \) is \( v = 6.00 \, \text{m/s} \)

We are asked to find the **amplitude** of the motion.

### Solution Outline
For a system undergoing simple harmonic motion (SHM), the **total mechanical energy** \( E \) is conserved. This energy consists of:
1. **Kinetic energy** at position \( x \): \( KE = \frac{1}{2}mv^2 \)
2. **Potential energy** in the spring at position \( x \): \( PE = \frac{1}{2}kx^2 \)

The total energy \( E \) in the SHM system is:
\[
E = KE + PE
\]

At the amplitude \( A \) (maximum displacement), all the energy is stored as potential energy:
\[
E = \frac{1}{2}kA^2
\]

### Step-by-Step Solution
1. **Calculate the kinetic energy** at \( x = 0.200 \, \text{m} \):
   \[
   KE = \frac{1}{2}mv^2 = \frac{1}{2} \cdot 0.400 \, \text{kg} \cdot (6.00 \, \text{m/s})^2
   \]
   \[
   KE = 0.200 \cdot 36 = 7.2 \, \text{J}
   \]

2. **Calculate the potential energy** at \( x = 0.200 \, \text{m} \):
   \[
   PE = \frac{1}{2}kx^2 = \frac{1}{2} \cdot 280 \, \text{N/m} \cdot (0.200 \, \text{m})^2
   \]
   \[
   PE = 140 \cdot 0.04 = 5.6 \, \text{J}
   \]

3. **Calculate the total energy** \( E \):
   \[
   E = KE + PE = 7.2 \, \text{J} + 5.6 \, \text{J} = 12.8 \, \text{J}
   \]

4. **Solve for the amplitude** \( A \) using \( E = \frac{1}{2}kA^2 \):
   \[
   12.8 = \frac{1}{2} \cdot 280 \cdot A^2
   \]
   \[
   A^2 = \frac{12.8 \times 2}{280} = \frac{25.6}{280} = 0.09143
   \]
   \[
   A = \sqrt{0.09143} \approx 0.302 \, \text{m}
   \]

### Answer:
The amplitude of the motion is approximately \( \mathbf{0.302 \, \text{m}} \).

So, the correct answer is:
**(c) 0.302 m**

Would you like more details or have any questions?

---

### Related Questions
1. How is total mechanical energy conserved in simple harmonic motion?
2. Why is the potential energy maximum at the amplitude in simple harmonic motion?
3. What role does the spring constant play in determining the amplitude?
4. How would the amplitude change if the speed at \( x = 0.200 \, \text{m} \) were doubled?
5. Can this method be used if there were friction on the surface?

### Tip:
In simple harmonic motion, understanding energy distribution (kinetic vs. potential) at different points can simplify solving for quantities like amplitude and maximum velocity.

A block with mass 0.400 kg is attached to a horizontal spring that has force constant k = 280 N/m. The block is moving in simple harmonic motion on a frictionless surface. When the block is at x = 0.200 m, it has speed 6.00 m/s. What is the amplitude of the motion of the block?

Learn how to calculate the amplitude of a block in simple harmonic motion attached to a horizontal spring with a force constant of 280 N/m. This solution involves concepts of energy conservation and simple harmonic motion, using formulas for kinetic and potential energy. Suitable for undergraduate physics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation