To solve this problem, we'll use Hooke's Law, which states that the force required to stretch or compress a spring is proportional to the distance stretched or compressed from its natural length. The work done in stretching a spring can be found using the formula:

$W = \frac{1}{2} k x^2$

where:

• $W$ is the work done,
• $k$ is the spring constant,
• $x$ is the displacement from the natural length.

Step 1: Find the spring constant $k$

We are given that 5 J of work is needed to stretch the spring from 24 cm to 40 cm. The displacement from the natural length is:

$x = 40 \, \text{cm} - 24 \, \text{cm} = 16 \, \text{cm} = 0.16 \, \text{m}$

Using the formula for work:

$5 = \frac{1}{2} k (0.16)^2$

Solving for $k$:

$5 = \frac{1}{2} k (0.0256)$

$5 = 0.0128k$

$k = \frac{5}{0.0128} = 390.625 \, \text{N/m}$

Step 2: Calculate the work needed to stretch the spring from 28 cm to 36 cm

Now, let's calculate the work needed to stretch the spring from 28 cm to 36 cm:

• The initial displacement from the natural length (28 cm):

$x_1 = 28 \, \text{cm} - 24 \, \text{cm} = 4 \, \text{cm} = 0.04 \, \text{m}$

• The final displacement from the natural length (36 cm):

$x_2 = 36 \, \text{cm} - 24 \, \text{cm} = 12 \, \text{cm} = 0.12 \, \text{m}$

The work done to stretch the spring from 28 cm to 36 cm is given by:

$W = \frac{1}{2} k (x_2^2 - x_1^2)$

Substituting the values:

$W = \frac{1}{2} \times 390.625 \times (0.12^2 - 0.04^2)$

$W = \frac{1}{2} \times 390.625 \times (0.0144 - 0.0016)$

$W = \frac{1}{2} \times 390.625 \times 0.0128$

$W = 390.625 \times 0.0064 = 2.5 \, \text{J}$

Thus, the work needed to stretch the spring from 28 cm to 36 cm is 2.50 J.

Would you like further details or have any questions?

Here are some related questions you might be interested in:

1. How does the spring constant $k$ affect the work done on the spring?
2. What is Hooke's Law, and how is it derived?
3. How would the work required change if the spring were compressed instead of stretched?
4. How does the energy stored in a spring relate to its potential energy?
5. What would happen if the spring is stretched beyond its elastic limit?

Tip: Remember, the work done on a spring depends on the square of the displacement, so small changes in displacement can result in significant changes in the work required!