To solve this problem, we'll use concepts from Hooke's law, Newton's second law of motion, and the work-energy theorem. Let's break the problem into parts and solve them one by one.

Part (a) What is the force constant of the spring?

According to Hooke's Law:

$F = k \cdot x$

Where:

• $F$ is the force exerted by the object (equal to its weight, $F = mg$),
• $k$ is the spring constant (force constant) we need to find,
• $x$ is the displacement or stretch of the spring (in meters).

Given:

• Mass of the object, $m = 2.40 \, \text{kg}$,
• Displacement $x = 2.56 \, \text{cm} = 0.0256 \, \text{m}$,
• Gravitational acceleration $g = 9.81 \, \text{m/s}^2$.

First, calculate the force $F$:

$F = m \cdot g = 2.40 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 23.544 \, \text{N}$

Now, use Hooke's law to find the spring constant $k$:

$k = \frac{F}{x} = \frac{23.544 \, \text{N}}{0.0256 \, \text{m}} = 919.69 \, \text{N/m}$

So, the force constant $k$ of the spring is approximately 919.7 N/m.

---

Part (b) How far will the spring stretch if a 1.20-kg block is hung on it?

We'll apply Hooke's law again, but now with the new mass of $1.20 \, \text{kg}$.

• Mass $m = 1.20 \, \text{kg}$,
• Force constant $k = 919.7 \, \text{N/m}$,
• Gravitational acceleration $g = 9.81 \, \text{m/s}^2$.

First, calculate the force exerted by the 1.20-kg block:

$F = m \cdot g = 1.20 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 11.772 \, \text{N}$

Now, find the stretch $x$ using Hooke's law:

$x = \frac{F}{k} = \frac{11.772 \, \text{N}}{919.7 \, \text{N/m}} = 0.0128 \, \text{m} = 1.28 \, \text{cm}$

So, the spring will stretch 1.28 cm when the 1.20-kg block is hung on it.

---

Part (c) How much work must an external agent do to stretch the same spring 9.00 cm from its unstretched position?

The work done to stretch a spring is given by the equation for the elastic potential energy stored in the spring:

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

Where:

• $k = 919.7 \, \text{N/m}$ (the spring constant),
• $x = 9.00 \, \text{cm} = 0.0900 \, \text{m}$ (the displacement).

Now, substitute the values into the equation:

$W = \frac{1}{2} \cdot 919.7 \, \text{N/m} \cdot (0.0900 \, \text{m})^2$

$W = \frac{1}{2} \cdot 919.7 \cdot 0.0081 \, \text{m}^2 = 3.724 \, \text{J}$

So, the work done by the external agent to stretch the spring 9.00 cm is approximately 3.72 joules.

---

Do you have any questions or want details on any part? Here are 5 related questions:

1. How does the spring constant affect the stretch of a spring for different masses?
2. What happens to the work required if the stretch of the spring is doubled?
3. Can the spring stretch indefinitely according to Hooke's law?
4. How would the solution change if the spring were not ideal (e.g., if it deformed permanently)?
5. What are some real-world applications of Hooke's law in engineering?

Tip: Remember, the work done on a spring is proportional to the square of the displacement. Therefore, doubling the stretch quadruples the required work!