Question

When an object of mass \(5 \mathrm{~kg}\) is suspended from a spring, the spring is observed to stretch by \(8 \mathrm{~cm}\). The deflection of the spring is related linearly to the weight of the suspended mass. What is the proportionality constant, in newton per \(\mathrm{cm}\), if \(g=\) \(9.81 \mathrm{~m} / \mathrm{s}^{2} ?\)

Short answer

The proportionality constant is \( 6.13125 \mathrm{~N/cm} \).

Step-by-step solution

Understand the Problem

Determine the relationship between the weight of the object and the spring's deflection. The deflection is linearly proportional to the weight.

Calculate the Weight of the Object

Use the formula to calculate the weight:\( W = mg \)where \( m = 5 \mathrm{~kg} \) and \( g = 9.81 \mathrm{~m/s^2} \). Hence, \( W = 5 \mathrm{~kg} \times 9.81 \mathrm{~m/s^2} = 49.05 \mathrm{~N} \).

Express the Deflection in Meters

Convert the deflection from cm to meters:\( 8 \mathrm{~cm} = 0.08 \mathrm{~m} \).

Calculate the Proportionality Constant

The proportionality constant \( k \) is defined as the ratio of the weight to the deflection:\[ k = \frac{W}{d} \]Substitute the known values:\[ k = \frac{49.05 \mathrm{~N}}{0.08 \mathrm{~m}} = 613.125 \mathrm{~N/m} \].

Convert the Proportionality Constant to N/cm

Since 1 m = 100 cm, convert \( k \) to N/cm:\[ k = 613.125 \mathrm{~N/m} \times \frac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 6.13125 \mathrm{~N/cm} \].

Key concepts

Mass and Weight Relationship

To understand the relationship between mass and weight, remember that weight is the force exerted by gravity on an object. It depends on the mass of the object and the gravitational acceleration. The formula to calculate weight (\textbf{W}) is:
\[ W = mg \]
Here, (\textbf{m}) is the mass in kilograms, and (\textbf{g}) is the acceleration due to gravity, typically \( 9.81 \text{m/s}^2\) on Earth.
For example, in the given exercise, the object’s mass is \( 5 \text{kg}\). Multiplying this by \( 9.81 \text{m/s}^2\), we find the weight (\textbf{W}) to be \( 49.05 \text{N}\).
This helps us see how mass and weight are directly related: the more mass an object has, the greater its weight.

Linear Deflection

Linear deflection refers to the stretching or compressing of a spring when a force is applied. This relationship is described by Hooke's Law, which states that the force (\textbf{F}) applied to a spring is directly proportional to the deflection or change in length (\textbf{d}) of the spring.
The formula is: \[ F = kd \]
Here, \( k \) is the spring constant, a measure of the stiffness of the spring.
In our exercise, when the 5 kg mass is suspended, it stretches the spring by 8 cm (\textbf{0.08 m}). This means the force due to weight (49.05 N) creates a deflection of 0.08 meters in the spring.
Linear deflection is crucial for understanding how materials respond to forces and is applied in many fields, including engineering and physics.

Proportionality Constant

The proportionality constant (\textbf{k}), often referred to as the spring constant, indicates how stiff a spring is. It quantifies the relationship between the force (\textbf{F}) applied to the spring and the resulting deflection (\textbf{d}).
The formula to determine the spring constant is: \[ k = \frac{F}{d} \]
In the exercise, we used the weight of the object (\textbf{49.05 N}) and the deflection (\textbf{0.08 m}) to find \( k \). Substituting the values,
\[ k = \frac{49.05 \text{N}}{0.08 \text{m}} = 613.125 \text{N/m} \]
Since 1 meter equals 100 centimeters, we convert \( k \) to N/cm by dividing by 100:
\[ k = 6.13125 \text{N/cm} \]
This tells us that for every centimeter, the spring exerts a force of about 6.13125 N. The spring constant is a fundamental property in designing systems that involve springs and in studying material properties.