We are asked to use Hooke's Law to determine how far a spring is stretched by a force of 50 N, given that 20 J of work is needed to stretch the spring from its natural length of 0.3 meters to a length of 0.4 meters.

Steps:

1. Work-energy relation for a spring is given by: $W = \frac{1}{2} k x^2$ where $W$ is the work, $k$ is the spring constant, and $x$ is the displacement from the natural length.

2. First, find the spring constant $k$ using the work $W = 20 \, \text{J}$ and the displacement from the natural length $x = 0.4 - 0.3 = 0.1 \, \text{m}$: $20 = \frac{1}{2} k (0.1)^2$ Solving this for $k$.

3. Once we find $k$, we will use Hooke's Law to determine the new displacement when a force of 50 N is applied: $F = k x$ where $F = 50 \, \text{N}$ and $x$ is the displacement we want to find.

Let's compute this.1. The spring constant $k$ is calculated to be 4000 N/m. 2. Using Hooke's Law with a force of 50 N, the spring will stretch an additional 0.0125 meters (or 1.25 cm) beyond its natural length.

Would you like further clarification or details on this?

Here are 5 related questions you might be interested in:

1. How is the spring constant $k$ related to the stiffness of the spring?
2. How would the displacement change if the force applied were doubled?
3. What if the work required to stretch the spring was different? How would that affect $k$?
4. How can Hooke’s Law be applied to non-linear springs?
5. How do real-life factors, like temperature, affect spring behavior?

Tip: The work done on a spring increases quadratically with the displacement, meaning doubling the displacement requires four times the work!