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The spring constant, denoted as \( k \), is a valuable factor in Hooke's Law. Understanding this constant helps measure the stiffness of a spring. It tells us how much force is required to stretch or compress the spring by a certain distance.To find the spring constant, we need the force \( F \) applied to the spring and the resulting displacement \( x \). Hooke's Law gives us the formula \( F = kx \). From this, we can solve for \( k \) using the equation \( k = \frac{F}{x} \).

In our exercise, a scale acting as a spring is compressed by a maximum load of 120 kg, which creates a force since weight is the force due to gravity (\( F = mg \)). With \( g = 9.81 \text{ m/s}^2\), the force calculation is straightforward: \( F = 120 \text{ kg} \times 9.81 \text{ m/s}^2 = 1177.2 \text{ N} \). With the spring displacement \( x = 0.75 \text{ cm} \) (converted to meters as 0.0075 m), we find the spring constant \( k = \frac{1177.2 \text{ N}}{0.0075 \text{ m}} = 157,280 \text{ N/m} \). This high value indicates a very stiff spring.

Calculating force is a crucial step in applying Hooke's Law. The force exerted by an object due to gravity is given by the equation \( F = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (typically \( g = 9.81 \text{ m/s}^2 \) on Earth).

For our problem:

• The rugby player's weight creates a force on the spring scale.
• His depression of the scale by 0.48 cm is the starting point for finding the force he exerts.

Convert 0.48 cm to meters, which is 0.0048 m. Plugging this into Hooke's Law where \( k = 157,280 \text{ N/m} \) (calculated earlier), we find the force: \( F_2 = kx_2 = 157,280 \text{ N/m} \times 0.0048 \text{ m} = 754.944 \text{ N} \). This force calculation is vital as it directly links the displacement on the spring scale to the player's weight.

In physics, unit conversion is essential for accurate calculation and interpretation of results. Distances in this problem are initially given in centimeters, and we must convert them into meters to match the standard unit in physics for distance (meters) in SI units.

Consider:

• The scale depression of 0.75 cm converts to 0.0075 m.
• The player's spring compression of 0.48 cm is 0.0048 m when converted.

Using consistent units allows Hooke's Law to be correctly applied, ensuring that the force (in Newtons) and displacement (in meters) are compatible.
This attention to unit conversion is a fundamental step, often leading to errors if overlooked. It's important to always verify the units when dealing with physical equations, enabling precise calculations and correct understanding of physical phenomena.