Problem 26 - Master Resource Book in JEE Main Physics

Question

A spring of spring constant \(5 \times 10^{3} \mathrm{Nm}^{-1}\) is stretched initially by \(5 \mathrm{~cm}\) from the unstretched position. Then the work required to stretch it further by another \(5 \mathrm{~cm}\) is (a) \(12.50 \mathrm{~N}-\mathrm{m}\) (b) \(18.75 \mathrm{~N}-\mathrm{m}\) (c) \(25.00 \mathrm{~N}-\mathrm{m}\) (d) \(6.25 \mathrm{~N}-\mathrm{m}\)

Step-by-Step Solution

Answer

The work required to stretch the spring further by another 5 cm is \(18.75 \mathrm{~N}-\mathrm{m}\), which is option (b).

Step 1: Understanding the Problem

We are given a spring with a spring constant, \( k = 5 \times 10^{3} \, \mathrm{Nm}^{-1} \), and we need to find the work required to stretch it further by \(5 \, \mathrm{cm}\) after it has already been stretched \(5 \, \mathrm{cm}\). To find this, we will use the formula for the work done in stretching a spring.

Step 2: Identifying the Work Formula

The work done in stretching or compressing a spring by a distance \( x \) from its natural length is given by the formula: \( W = \frac{1}{2} k x^2 \). This formula calculates the energy (work) required for maintaining a change in the deformation of a spring.

Step 3: Initial Work Calculation

First, calculate the work done to stretch the spring by the initial \( 5 \, \mathrm{cm} \), which is \( 0.05 \, \mathrm{m} \). Use the formula: \[ W_1 = \frac{1}{2} \times 5 \times 10^{3} \times (0.05)^2 \] Calculate this value: \( W_1 = \frac{1}{2} \times 5000 \times 0.0025 = 6.25 \mathrm{~N}-\mathrm{m} \).

Step 4: Total Work Calculation After Further Stretch

Next, find the work required to stretch the spring from the initial \( 5 \, \mathrm{cm} \) to \( 10 \, \mathrm{cm} \), which is a total of \( 0.10 \, \mathrm{m} \). This total work is \[ W_2 = \frac{1}{2} \times 5 \times 10^{3} \times (0.10)^2 \] Calculate: \( W_2 = \frac{1}{2} \times 5000 \times 0.01 = 25.00 \mathrm{~N}-\mathrm{m} \).

Step 5: Calculating Additional Work

The additional work required to stretch the spring further by another \( 5 \, \mathrm{cm} \) is the difference between the work calculated for \( 10 \, \mathrm{cm} \) and \( 5 \, \mathrm{cm} \): \[ \Delta W = W_2 - W_1 \] Thus, \( \Delta W = 25.00 \mathrm{~N}-\mathrm{m} - 6.25 \mathrm{~N}-\mathrm{m} = 18.75 \mathrm{~N}-\mathrm{m} \).

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle that describes the behavior of springs. It states that the force required to extend or compress a spring is directly proportional to the distance it is stretched or compressed from its rest position. Mathematically, it's expressed as \( F = kx \, where: \ F \) is the force applied to the spring, \ \( k \) is the spring constant, and \ \( x \) is the displacement from the equilibrium position.
This simple linear relationship holds true as long as the elastic limit of the material is not exceeded. Hooke's Law gives us a way to predict how much a spring will stretch or compress for a given force, making it crucial in physics and engineering for tasks ranging from designing structures to creating sensitive measuring devices. Understanding this law allows you to calculate the potential energy stored in the spring, along with other related dynamics.

Spring Constant

The spring constant, symbolized by \( k \), is a measure of a spring's stiffness. It defines the relationship between the force applied to a spring and the resulting amount of stretch or compression. A large spring constant means the spring is stiff, requiring more force to stretch. Conversely, a small spring constant indicates a less stiff spring that stretches more easily.
The spring constant is an intrinsic property of the spring determined by its material properties and geometry. It is measured in units of force per unit length, typically Newtons per meter (\( \mathrm{Nm}^{-1} \)). Knowing the spring constant helps you calculate the amount of force needed to achieve a certain deformation and is crucial for tasks like controlling mechanical vibrations, understanding balance controls, or building systems requiring precise motion like watches or test stands.

Elastic Potential Energy

Elastic potential energy is the energy stored in elastic materials as a result of their stretching or compressing. For springs, this energy is governed by Hooke's Law and is given by the formula: \[ E_{pe} = \frac{1}{2} k x^2 \] \, where \( E_{pe} \) is the elastic potential energy, \( k \) is the spring constant, and \( x \) is the displacement from the spring's equilibrium position.
When a spring is stretched or compressed, it holds energy that can be released when it returns to its original (equilibrium) state. Elastic potential energy is vital in various applications, including vehicle suspensions that provide a comfortable ride and gymnastics equipment that supports athletic performances. Knowing how to calculate this energy helps in designing systems that make efficient use of energy storage and release, like mechanisms for launching projectiles or improvements in renewable energy systems like wave or wind power.