A body of mass m=3.90 kg slides on a horizontal frictionless table with a speed of v=120 `m s^(-1)`. It is brought to rest in compressing a spring in its path. How much does spring is compressed, if its force constant k is 135 `Nm^(-1)`?

Options

A. 0.204 m
B. 0.408 m
C. 0.804 m
D. 4.04 m

Explanation

To solve the problem, we need to find out how much the spring is compressed when a body of mass \( m = 3.90 \, \text{kg} \) sliding with a speed \( v = 1.20 \, \text{m/s} \) comes to rest after compressing the spring with a spring constant \( k = 135 \, \text{N/m} \).

### Step-by-Step Solution:

1. **Identify the Energy Types**:
- The body has kinetic energy (KE) when it is moving.
- When the body compresses the spring, the kinetic energy is converted into potential energy (PE) stored in the spring.

2. **Write the Kinetic Energy Formula**:
\[ KE = \frac{1}{2} m v^2 \]
where:
- \( m = 3.90 \, \text{kg} \)
- \( v = 1.20 \, \text{m/s} \)

3. **Calculate the Kinetic Energy**:
\[ KE = \frac{1}{2} \times 3.90 \times (1.20)^2 \]
\[ KE = \frac{1}{2} \times 3.90 \times 1.44 \]
\[ KE = 2.7952 \, \text{J} \]

4. **Write the Potential Energy Formula for the Spring**:
\[ PE = \frac{1}{2} k x^2 \]
where \( x \) is the compression of the spring.

5. **Set the Kinetic Energy Equal to the Potential Energy**:
Since all the kinetic energy is converted into potential energy:
\[ \frac{1}{2} m v^2 = \frac{1}{2} k x^2 \]

6. **Cancel Out the \(\frac{1}{2}\)**:
\[ m v^2 = k x^2 \]

7. **Rearrange to Solve for \( x^2 \)**:
\[ x^2 = \frac{m v^2}{k} \]

8. **Substitute the Known Values**:
\[ x^2 = \frac{3.90 \times (1.20)^2}{135} \]
\[ x^2 = \frac{3.90 \times 1.44}{135} \]
\[ x^2 = \frac{5.616}{135} \]
\[ x^2 \approx 0.0417 \]

9. **Take the Square Root to Find \( x \)**:
\[ x = \sqrt{0.0417} \approx 0.204 \, \text{m} \]

### Final Answer:
The spring is compressed by approximately \( x \approx 0.204 \, \text{m} \).