if 32 joules of work is needed to stretch a spring 40 cm beyond equilibrium, how much work is required to compress it 6 m beyond equilibrium

The work required to either stretch or compress a spring beyond its equilibrium is governed by **Hooke's Law** and the formula for the potential energy stored in a spring:

\[
W = \frac{1}{2} k x^2
\]

Where:
- \(W\) is the work done on the spring.
- \(k\) is the spring constant.
- \(x\) is the displacement from the equilibrium position.

### Step 1: Use the given information to find the spring constant, \(k\).
We know:
- \(W = 32 \, \text{J}\) (work to stretch the spring),
- \(x = 40 \, \text{cm} = 0.40 \, \text{m}\).

Substitute into the work formula:

\[
32 = \frac{1}{2} k (0.40)^2
\]

Solve for \(k\):

\[
32 = \frac{1}{2} k (0.16)
\]

\[
32 = 0.08 k
\]

\[
k = \frac{32}{0.08} = 400 \, \text{N/m}
\]

### Step 2: Calculate the work required to compress the spring by 6 meters.
Now we need to find the work when \(x = 6 \, \text{m}\) and \(k = 400 \, \text{N/m}\).

\[
W = \frac{1}{2} k x^2 = \frac{1}{2} \times 400 \times (6)^2
\]

\[
W = 200 \times 36 = 7200 \, \text{J}
\]

### Conclusion:
The work required to compress the spring by 6 meters is **7200 joules**.

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Would you like further details on any of the steps?

Here are 5 related questions to explore further:
1. How does Hooke's Law apply to springs beyond equilibrium?
2. What happens when the displacement is in the opposite direction (compression)?
3. How does the spring constant relate to the stiffness of a spring?
4. How does the work change with different spring constants?
5. What happens to the potential energy when the displacement doubles?

**Tip:** The work done on a spring increases quadratically with displacement, meaning small changes in displacement can result in large changes in required work.

If 32 joules of work is needed to stretch a spring 40 cm beyond equilibrium, how much work is required to compress it 6 m beyond equilibrium?

Learn how to calculate the work required to stretch or compress a spring using Hooke's Law. In this example, 32 joules of work is used to stretch a spring 40 cm, and we calculate the work to compress it by 6 meters. This problem includes key physics concepts such as spring constant and energy stored in springs. Suitable for students in Grades 10-12.

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