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In this problem, we are given data about a spring being stretched by different weights, and the task is to find a linear model that describes the length of the spring as a function of the mass of the weight. Let's go through the parts step-by-step:

### Given Data:

| Mass \( m \) (grams) | 40  | 60  | 100  | 200  | 500  |
|----------------------|-----|-----|------|------|------|
| Length \( L \) (cm)   | 15.5| 19.8| 24.1 | 31.7 | 51.9 |

#### Task (a):
We are asked to determine a linear model of the form:

\[
L(m) = \alpha \cdot m + \beta
\]

Where \( L \) is the length of the spring (in cm), \( m \) is the mass of the weight (in grams), and \( \alpha \) and \( \beta \) are constants to be determined from the data.

#### Approach:
1. We need to use linear regression or the least squares method to fit the best line to the given points.
2. This involves solving for \( \alpha \) (slope) and \( \beta \) (intercept).

#### Task (b):
After finding the linear equation, we are asked to explain the meaning of \( \alpha \), which represents how much the length of the spring increases per unit mass, i.e., the "spring constant" in a simplified form.

I will now calculate the linear model and return with the result.The linear model for the length \( L(m) \) of the spring as a function of the mass \( m \) is:

\[
L(m) = 0.0752 \cdot m + 15.065
\]

### Explanation:
- The slope \( \alpha = 0.0752 \) indicates that for every additional gram of mass added to the spring, the length of the spring increases by approximately 0.0752 cm.
- The intercept \( \beta = 15.065 \) represents the initial length of the spring when no mass is applied (assuming \( m = 0 \), though not directly measured in the table).
- The coefficient of determination \( R^2 = 0.985 \), which is very close to 1, suggests that this linear model fits the data extremely well.

#### Task (b):
In this context, the constant \( \alpha = 0.0752 \) can be interpreted as the rate at which the spring stretches per unit of mass. It is related to the stiffness of the spring: a smaller value of \( \alpha \) would indicate a stiffer spring, and a larger value would indicate a more elastic spring.

Would you like any more details or further clarification on these points?

Here are five additional questions to deepen your understanding:
1. How would the model change if we used a nonlinear fit, such as a quadratic?
2. What would the physical interpretation of the intercept \( \beta \) be in real-world terms?
3. Can you predict the length of the spring for a mass of 300 grams using the model?
4. How would the model change if you included more data points for smaller masses?
5. Could you explore how changing the spring material might affect the slope \( \alpha \)?

**Tip:** Understanding \( R^2 \) can help assess the accuracy of linear models. The closer it is to 1, the better the fit.

Opgave 4: En fjeder belastes med forskellige lodder i fysiklokalet. I tabel 1 ses sammenhørende målte værdier for massen af loddet og fjederens længde. Massen måles i gram, og længden af fjederen i centimeter. a) Bestem en lineær model L(x) = αx + b, der beskriver længden af fjederen som funktion af massen af loddet, og begrund hvor godt denne model passer ud fra målingerne. b) Forklar betydningen af konstanten α i denne situation.

Solve a physics problem where the length of a spring is modeled using a linear equation as a function of mass. Learn how to derive the model L(m) = 0.0752 * m + 15.065 and understand the meaning of the constant α, which represents how much the spring stretches per unit mass.

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