Problem 47 - Master Resource Book in JEE Main Physics

Question

Assertion Two springs of force constants \(k_{1}\) and \(k_{2}\) are stretched by the same force. If \(k_{1}>k_{2}\), then work done in stretching the first \(\left(W_{1}\right)\) is less than work done in stretching the second \(\left(W_{2}\right.\) ). Reason \(F=k_{1} x_{1}=k_{2} x_{2}\) $$ \begin{aligned} \frac{x_{1}}{x_{2}} &=\frac{k_{2}}{k_{1}} \\ \frac{W_{1}}{W_{2}} &=\frac{\frac{1}{2} k_{1} x_{1}^{2}}{\frac{1}{2} k_{2} x_{2}^{2}}=\frac{k_{1}}{k_{2}}\left(\frac{k_{2}}{k_{1}}\right)^{2}=\frac{k_{2}}{k_{1}} \end{aligned} $$ As \(k_{1}>k_{2}, W_{1}

Step-by-Step Solution

Answer

Yes, \(W_1 < W_2\) because \(k_1 > k_2\).

Step 1: Understanding the Problem

We have two springs with force constants (spring constants) \(k_1\) and \(k_2\), where \(k_1 > k_2\). Both springs are stretched by the same force \(F\). We need to compare the work done in stretching each spring, denoted as \(W_1\) for the spring with constant \(k_1\) and \(W_2\) for the spring with constant \(k_2\).

Step 2: Recall Hooke's Law

According to Hooke's Law, the force required to stretch or compress a spring by some distance \(x\) is given by \(F = kx\). For the two springs, these equations give us \(F = k_1 x_1\) and \(F = k_2 x_2\), where \(x_1\) and \(x_2\) are the extensions or compressions in the respective springs.

Step 3: Establish the Relationship Between Extensions

From the Hooke's Law equations, since \(F = F\text{ (the same for both springs)},\), we equate: \(k_1 x_1 = k_2 x_2\). Solving for the ratio of extensions gives: \(\frac{x_1}{x_2} = \frac{k_2}{k_1}\). This means the spring with the larger constant (\(k_1\)) stretches less.

Step 4: Calculate Work Done on Each Spring

The work done \(W\) in stretching a spring is given by: \(W = \frac{1}{2}kx^2\). Thus, we have: \(W_1 = \frac{1}{2} k_1 x_1^2\) and \(W_2 = \frac{1}{2} k_2 x_2^2\).

Step 5: Compare the Work Done

We substitute \(x_1 = \frac{k_2}{k_1} x_2\) into \(W_1\):\[\begin{align*} W_1 &= \frac{1}{2} k_1 \left(\frac{k_2}{k_1} x_2\right)^2 \ &= \frac{1}{2} k_1 \frac{k_2^2}{k_1^2} x_2^2 \ &= \frac{k_2}{k_1} \frac{1}{2} k_2 x_2^2 \ &= \frac{k_2}{k_1} W_2 . \end{align*} \] Since \(k_1 > k_2\), it follows that \(\frac{k_2}{k_1} < 1\), which implies \(W_1 < W_2\).

Key Concepts

Hooke's Law

In the realm of mechanics, Hooke's Law serves as a fundamental principle describing how materials, specifically springs, behave when they are subjected to stretching or compressing forces. Simply put, it relates the force required to stretch or compress a spring to the distance the spring moves. Hooke's Law is mathematically expressed as:\[ F = kx \]Here, \( F \) stands for the force applied to the spring, \( k \) is the spring constant (an intrinsic property of each spring), and \( x \) represents the extension or compression distance of the spring. This relationship means that the force is directly proportional to the distance stretched or compressed—doubling the distance requires double the force.Hooke's Law applies only within a certain limit, known as the elastic limit, beyond which the spring may be deformed permanently and not return to its original shape.

• It is crucial in calculating how much force is needed to change the length of a spring.
• This law forms the basis for understanding spring dynamics in mechanics.
• Important for designing systems involving oscillations or vibrations such as shock absorbers.

Understanding this relationship allows us to explore more complex spring mechanics, such as the work done in stretching a spring.

Spring Constant

The spring constant, denoted by \( k \), is one of the key elements in understanding spring mechanics. It quantifies the stiffness of a spring. A higher spring constant indicates a stiffer spring that requires more force to stretch or compress the same distance as a spring with a lower constant.The spring constant is measured in newtons per meter (N/m) and is unique to every spring based on its material and geometry. To put it practically:- A large spring constant means the spring is stiff. It's harder to stretch or compress, which is common in heavy-duty applications.- A small spring constant indicates a softer spring, easier to manipulate. Such springs are used in devices requiring gentle forces.The spring constant is crucial when determining the behavior of a spring system under load. In our given exercise, comparing two springs with different constants helps illustrate their respective extents and the work done on them when a similar force is applied.Utilizing the spring constant in equations allows engineers and physicists to predict how a spring will behave in a particular scenario, offering a standardized way to compare different springs.

Mechanics

Mechanics is the branch of physics concerned with the behavior of objects under forces. It encompasses the study of motion and the application of forces to achieve it. Within this field, analyzing springs under applied forces represents a typical study of elasticity and dynamics. Studying mechanics allows us to understand many aspects of real-world applications, such as how a vehicle's suspension system mitigates road shocks using springs. By understanding the principles of mechanics, we can analyze both static and dynamic systems to predict their behavior under specific conditions. Key areas of study in mechanics include:

• Statics: Examining bodies at rest, or in equilibrium.
• Dynamics: Studying bodies in motion and the forces causing that motion.
• Kinetics and kinematics: Examining the motion of bodies and how forces affect them.

In the context of springs, mechanics involves understanding how forces are transferred through the spring and how the spring's properties, like the spring constant, influence its interaction with other objects.