Question

A spring of spring constant \(10^{3} \mathrm{~N} / \mathrm{m}\) is stretched initially \(4 \mathrm{~cm}\) from the unscratched position. How much the work required to stretched it further by another \(5 \mathrm{~cm}\) ? (A) \(6.5 \mathrm{NM}\) (B) \(2.5 \mathrm{NM}\) (C) \(3.25 \mathrm{NM}\) (D) \(6.75 \mathrm{NM}\)

Short answer

The short answer for this question is: The amount of work required to stretch the spring further by another \(5 \mathrm{~cm}\) is approximately \(3.25 \mathrm{~N \cdot m}\), which corresponds to option (C).

Step-by-step solution

Identify the known values

We are given the spring constant \(k = 10^3 \mathrm{~N/m}\), the initial stretched length \(x_{1} = 4 \mathrm{~cm}\), and the length to be stretched further (\(5 \mathrm{~cm}\)). We need to find the work done (W) to stretch the spring by the given length.

Convert length to meters

The given lengths are in centimeters, and we need to convert them to meters for our calculations, since the spring constant is given in \(\mathrm{N/m}\). So, \(x_{1} = 4\mathrm{~cm} = 0.04\mathrm{~m}\) The length to be stretched further is \(5\mathrm{~cm} = 0.05\mathrm{~m}\)

Calculate the final stretched length

We need to find the final stretched length by adding the initial stretched length and the length to be stretched further. \(x_{2} = x_{1} + 0.05 \mathrm{~m} = 0.04\mathrm{~m} + 0.05\mathrm{~m} = 0.09\mathrm{~m}\)

Calculate the work done

Now, we can plug in the values into the formula of work done by the spring, which is given by Hooke's law: $$W = \frac{1}{2}k(x_{2}^2 - x_{1}^2)$$ Replacing the known values: $$W = \frac{1}{2}(10^3 \mathrm{~N/m})(0.09^2\mathrm{~m^2} - 0.04^2\mathrm{~m^2})$$ Calculating the work done: $$W \approx 3.25 \mathrm{~N\cdot m}$$

Compare the result with the given options

We calculated the work done to be approximately \(3.25 \mathrm{~N\cdot m}\). Comparing this result with the given options, we see that the closest answer is: (C) \(3.25 \mathrm{~N \cdot m}\)

Key concepts

Spring Constant

The spring constant, denoted as \( k \), is a measure of a spring's stiffness. It is defined as the force required to compress or stretch the spring by a unit length.
In the exercise, the spring constant is given as \( 10^3 \text{ N/m} \). This means for every meter we stretch or compress the spring, a force of \( 1000 \text{ N} \) is needed.
Springs with a higher spring constant value are stiffer and require more force to deform, while springs with a lower spring constant are more easily compressed or stretched. Understanding the spring constant is crucial when calculating forces and work involving springs.

Hooke's Law

Hooke's Law is a fundamental principle that relates the force applied to a spring and the resulting displacement. Formally, it is written as \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the spring's equilibrium position.
This law shows that the force required to stretch or compress a spring is directly proportional to the displacement. It is essential for determining how much work is done when deforming a spring.
When dealing with energy in springs, Hooke's Law helps us find the work done using the formula \( W = \frac{1}{2} k x^2 \). This formula tells us the amount of energy stored in a spring when it is moved from its natural length.

Energy in Springs

Energy in springs is a crucial concept, particularly in mechanical systems. The potential energy stored in a spring, commonly known as elastic potential energy, is quantifiable using the formula \( W = \frac{1}{2} k (x_2^2 - x_1^2) \).
Where \( x_2 \) and \( x_1 \) are the final and initial displacements, respectively, and \( k \) is the spring constant. This formula is derived from Hooke's Law.
When you compress or stretch a spring from its natural length, it stores energy, which can be released to do work, such as propelling an object or returning to its original shape. Understanding energy in springs is essential when calculating how much work needs to be done or how much energy will be released.

Unit Conversion

Unit conversion is a vital step in physics problems, especially when dealing with SI units. In the given exercise, we converted centimeters to meters because the spring constant was given in \( \text{N/m} \).
To convert from centimeters to meters, we simply divide by 100, as there are 100 centimeters in a meter. For example, \( 4 \text{ cm} = 0.04 \text{ m} \) and \( 5 \text{ cm} = 0.05 \text{ m} \).
Correct unit conversions ensure the accuracy of calculations and allow for the consistent application of formulas. This is particularly important in physics, where mismatched units can lead to incorrect results.