To solve for the spring constant $k$ in this problem, we need to use the concept of work done in stretching a spring. The work $W$ required to stretch a spring is given by the integral of the force over the distance stretched.

The force required to stretch the spring by a distance $x$ from its natural length is given by Hooke's Law:

$f(x) = kx$

The work done in stretching the spring from its natural length to a distance $x$ is given by:

$W = \int_0^x f(x) \, dx = \int_0^x kx \, dx$

Now, evaluating this integral:

$W = k \int_0^x x \, dx = k \left[ \frac{x^2}{2} \right]_0^x = \frac{kx^2}{2}$

This is the expression for the work done to stretch the spring from its natural length to a length $x$.

Given Information:

• The spring is stretched from 36 cm to 56 cm, so the distance stretched is:

$x = 56 \, \text{cm} - 36 \, \text{cm} = 20 \, \text{cm} = 0.20 \, \text{m}$

• The work required to stretch the spring by this distance is 3 J:

$W = 3 \, \text{J}$

Using the work formula:

$3 = \frac{k(0.20)^2}{2}$

Now, solve for $k$:

$3 = \frac{k(0.04)}{2}$

$3 = 0.02k$

$k = \frac{3}{0.02} = 150 \, \text{N/m}$

Thus, the exact value of the spring constant $k$ is $150 \, \text{N/m}$.

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Related Questions:

1. How can you derive the work done formula for a spring using energy principles?
2. What is the significance of the spring constant $k$ in Hooke's Law?
3. How do you calculate the potential energy stored in a spring?
4. How does the work required to stretch a spring change if the spring constant is doubled?
5. Can Hooke's Law be applied to materials other than springs, like rubber bands?

Tip: The work done in stretching or compressing a spring is a form of potential energy stored in the spring.