Question

"A spring of natural length 1.5cm is extended by 0.02cm by a force of 1.5N.what will the length be when the applied force is 4.0N"

Answer 1

To determine the length of the spring when a force of 4.0 N is applied, we can use Hooke's Law, which states that the extension of a spring is directly proportional to the force applied to it, within the elastic limit of the material. Hooke's Law can be expressed as:

\[ F = k \cdot x \]

where:
- \( F \) is the force applied,
- \( k \) is the spring constant,
- \( x \) is the extension of the spring from its natural length.

1. First, we need to calculate the spring constant \( k \) using the initial force and extension:

Given:
- Natural length of the spring = 1.5 cm
- Extension at 1.5N = 0.02 cm

Extension in meters:
\[ x_1 = 0.02 \, \text{cm} = 0.0002 \, \text{m} \]

Using Hooke's Law:
\[ F_1 = k \cdot x_1 \]
\[ 1.5 = k \cdot 0.0002 \]

Solving for \( k \):
\[ k = \frac{1.5}{0.0002} = 7500 \, \text{N/m} \]

2. Now, we can find the extension when a force of 4.0 N is applied:

Given:
- \( F_2 = 4.0 \, \text{N} \)

Using Hooke's Law again:
\[ F_2 = k \cdot x_2 \]
\[ 4.0 = 7500 \cdot x_2 \]

Solving for \( x_2 \):
\[ x_2 = \frac{4.0}{7500} = 0.00053333 \, \text{m} \]
\[ x_2 \approx 0.0533 \, \text{cm} \]

3. Finally, we can find the new length of the spring when the force of 4.0 N is applied:

New length \( L \) is the natural length plus the extension:
\[ L = 1.5 \, \text{cm} + 0.0533 \, \text{cm} \]
\[ L \approx 1.5533 \, \text{cm} \]

So the length of the spring when a force of 4.0 N is applied will be approximately **1.553 cm**.