An ideal spring with spring constant K = 200 N / m is fixed on one end on a wall. If the spring is pulled with a force 10 N at the other end along its length, how much it will extended

Options

A. 5 cm
B. 2 m
C. 2 cm
D. 5 m

Explanation

To solve the problem of how much the spring will extend when a force of 10 N is applied, we can use Hooke's Law, which states that the force exerted by a spring is proportional to its extension or compression. The formula for Hooke's Law is:

\[ F = k \cdot x \]

where:
- \( F \) is the force applied to the spring (in Newtons),
- \( k \) is the spring constant (in N/m),
- \( x \) is the extension of the spring (in meters).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- The spring constant \( k = 200 \, \text{N/m} \)
- The applied force \( F = 10 \, \text{N} \)

2. **Apply Hooke's Law**:
According to Hooke's Law, we can rearrange the formula to solve for \( x \):
\[ x = \frac{F}{k} \]

3. **Substitute the Values**:
Now, substitute the known values into the equation:
\[ x = \frac{10 \, \text{N}}{200 \, \text{N/m}} \]

4. **Calculate the Extension**:
Perform the calculation:
\[ x = \frac{10}{200} = 0.05 \, \text{m} \]

5. **Convert to Centimeters**:
Since the question may require the answer in centimeters, convert meters to centimeters:
\[ x = 0.05 \, \text{m} \times 100 \, \text{cm/m} = 5 \, \text{cm} \]

### Final Answer:
The spring will extend by **5 centimeters** when a force of 10 N is applied.

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