Hooke's law states that the force needed to keep a spring stretched x units beyond its natural length is directly proportional to x. if a spring has a natural length of 10 cm, and a force of 40 N (newtons) is required to keep the spring stretched to a length to 15 cm , what force , in newtons, will be needed to keep the spring stretched to a length of 14 cm ?

Options

A. 8
B. 32
C. 37.3
D. 50

Answer

To solve the problem using Hooke's Law, we will follow these steps:

### Step 1: Identify the natural length and the stretched lengths
The natural length of the spring is given as 10 cm. We need to find the stretch when the spring is stretched to 15 cm and 14 cm.

### Step 2: Calculate the stretch for 15 cm
The stretch \( x_1 \) when the spring is stretched to 15 cm is calculated as:
\[ x_1 = 15 \, \text{cm} - 10 \, \text{cm} = 5 \, \text{cm} \]

### Step 3: Identify the force for the stretch of 5 cm
The force \( f_1 \) required to stretch the spring to 15 cm is given as 40 N.

### Step 4: Calculate the stretch for 14 cm
Now, we calculate the stretch \( x_2 \) when the spring is stretched to 14 cm:
\[ x_2 = 14 \, \text{cm} - 10 \, \text{cm} = 4 \, \text{cm} \]

### Step 5: Set up the proportion using Hooke's Law
According to Hooke's Law, the force is directly proportional to the stretch. Therefore, we can set up the proportion:
\[ \frac{f_1}{x_1} = \frac{f_2}{x_2} \]
Substituting the known values:
\[ \frac{40 \, \text{N}}{5 \, \text{cm}} = \frac{f_2}{4 \, \text{cm}} \]

### Step 6: Solve for \( f_2 \)
Cross-multiplying gives:
\[ 40 \, \text{N} \cdot 4 \, \text{cm} = f_2 \cdot 5 \, \text{cm} \]
\[ 160 \, \text{N} \cdot \text{cm} = f_2 \cdot 5 \, \text{cm} \]
Now, divide both sides by 5 cm:
\[ f_2 = \frac{160 \, \text{N} \cdot \text{cm}}{5 \, \text{cm}} = 32 \, \text{N} \]

### Conclusion
The force needed to keep the spring stretched to a length of 14 cm is \( \boxed{32} \) N.

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