Problem 436 - JEE Physics

Question

Natural length of a spring is \(60 \mathrm{~cm}\), and its spring constant is \(2000 \mathrm{~N} / \mathrm{m}\). A mass of \(20 \mathrm{~kg}\) is hung from it. The extension produced in the spring is..... \(\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(4.9 \mathrm{~cm}\) (B) \(0.49 \mathrm{~cm}\) (C) \(9.8 \mathrm{~cm}\) (D) \(0.98 \mathrm{~cm}\)

Step-by-Step Solution

Answer

The extension produced in the spring is (C) \(9.8 \mathrm{~cm}\).

Step 1: Convert units

First, it's important to have all units in the same system. Here, we need to convert the natural length of the spring from centimeters to meters. Natural length: \(60 \mathrm{~cm} = 0.6 \mathrm{~m}\)

Step 2: Calculate the force due to gravity on the mass

To find the force acting on the spring due to gravity, we use the formula: Force (\(F\)) = mass (\(m\)) * gravity (\(g\)) Given mass: \(20 \mathrm{~kg}\) and gravity: \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) The force acting on the spring due to mass: \(F = 20 \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2} = 196 \mathrm{~N}\)

Step 3: Apply Hooke's Law to find the extension

Hooke's Law states that the force acting on a spring is directly proportional to the extension produced in the spring: \(F = k \times x\) Where: - \(F\) is the force acting on the spring (\(196 \mathrm{~N}\)) - \(k\) is the spring constant (\(2000 \mathrm{~N} / \mathrm{m}\)) - \(x\) is the extension produced in the spring (the quantity we need to find) Now, rearrange the equation to solve for the extension \(x\): \(x = \frac{F}{k}\) Plug in the given values of the force and spring constant: \(x = \frac{196 \mathrm{~N}}{2000 \mathrm{~N} / \mathrm{m}}\)

Step 4: Solve for the extension

Perform the division to find the extension: \(x = 0.098 \mathrm{~m}\) Now, convert the extension back to centimeters: Extension: \(0.098 \mathrm{~m} = 9.8 \mathrm{~cm}\)

Step 5: Match the answer to the given options

The calculated extension is \(9.8 \mathrm{~cm}\), which matches option (C). Therefore, the correct answer is: (C) \(9.8 \mathrm{~cm}\)

Key Concepts

Spring Constant

The spring constant, often symbolized as "k," is a measure of a spring's stiffness. It tells us how much force is needed to stretch or compress a spring by a certain length. We measure this value in Newtons per meter ( / m"). A higher spring constant means the spring is stiffer, requiring more force to achieve the same amount of extension. So, if you think of a very rigid steel spring, it would have a larger spring constant compared to a soft rubber band.
In our exercise, the spring constant is 2000 N/m, which is pretty high, indicating the spring is relatively stiff. This explains why, even with a considerable force applied—196 N in our case—the extension is just a small distance. Understanding the spring constant is fundamental because it directly affects how stretchy or stiff materials behave when a force is applied on them.

Force Due to Gravity

The force of gravity acting on an object is calculated using the formula: \( F = m \times g \). Here, "m" represents mass, measured in kilograms, and "g" is the acceleration due to gravity, which is approximately 9.8 m/s² on Earth. This force is what pulls us and all objects towards the ground.
In the given problem, the mass of the hanging object is 20 kg. By plugging the values into the formula, the force exerted by gravity on this mass is calculated to be \( 196 \text{ N} \) (Newtons). This force is crucial as it directly impacts how the spring behaves when the weight is attached. Simply put, understanding this concept helps us calculate how much force is applied to any object due to Earth's gravitational pull, and consequently, how this force interacts with structures like springs.

Extension of Spring

When a force is applied to a spring, it stretches or compresses. This change in length is known as the extension of the spring. According to Hooke's Law, the extension is directly related to the force applied, represented by the equation \( F = k \times x \). Here, "F" is the force acting on the spring, "k" is the spring constant, and "x" is the extension.
In the exercise, the extension is the distance the spring stretches due to the mass of 20 kg attached to it. By rearranging Hooke's Law, \( x = \frac{F}{k} \), we can calculate this extension. Substituting the values we have—\( F = 196 \text{ N} \) and \( k = 2000 \text{ N/m} \)—we find \( x \approx 0.098 \text{ m} \), which converts to 9.8 cm. This result gives us the extension due to the weight hung from the spring.