To solve these problems, we use Hooke's Law and the work integral:

Step 1: Find the Spring Constant $k$

Hooke's Law states that the force required to compress or stretch a spring is:

$F = kx$

where:

• $F$ is the force applied,
• $k$ is the spring constant,
• $x$ is the displacement from the natural length.

Given:

• The natural length of the spring is 40 cm.
• A 60 N force is required to compress it by 10 cm.

Thus,

$60 = k(10)$

$k = \frac{60}{10} = 6 \text{ N/cm}$

or in meters:

$k = 600 \text{ N/m}$

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Step 2: Calculate Work Done for 10 cm Compression

The work done to compress a spring is given by the integral:

$W = \int_{x_1}^{x_2} kx \, dx$

For compression from $x = 0$ to $x = 10$ cm (0.1 m):

$W = \int_0^{0.1} 600x \, dx$

Solving:

$W = 600 \left[ \frac{x^2}{2} \right]_0^{0.1}$

$W = 600 \times \frac{(0.1)^2}{2}$

$W = 600 \times \frac{0.01}{2}$

$W = 600 \times 0.005 = 3 \text{ Joules}$

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Step 3: Calculate Work Done for Compression to 25 cm

A length of 25 cm means the displacement is:

$x = 40 - 25 = 15 \text{ cm} = 0.15 \text{ m}$

Now, we calculate:

$W = \int_0^{0.15} 600x \, dx$

$W = 600 \left[ \frac{x^2}{2} \right]_0^{0.15}$

$W = 600 \times \frac{(0.15)^2}{2}$

$W = 600 \times \frac{0.0225}{2}$

$W = 600 \times 0.01125$

$W = 6.75 \text{ Joules}$

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Final Answers:

• Work required for 10 cm compression: 3 Joules.
• Work required for 15 cm compression (to 25 cm length): 6.75 Joules.

Would you like a more detailed explanation of any step?

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Related Questions

1. How much work is required to stretch the spring by 15 cm instead of compressing it?
2. If we double the spring constant, how much work would be required for the same compression?
3. What is the force required to compress the spring to 25 cm?
4. What would be the natural frequency of oscillation for a mass attached to this spring?
5. How would the work required change if the spring obeyed a nonlinear force law?

Tip: The work done in compressing a spring follows a quadratic relationship with displacement, meaning doubling the compression distance quadruples the required work!