A lode attached to the end of a spring produces 9cm extension of spring and now if spring is cut into 3 equal parts and are connected paralleley to the same load then extension in the spring is?

To tackle this problem, we first need to understand how springs behave under load, particularly when it comes to extensions and how they interact when arranged in parallel. Let's break this down step by step.

Understanding Spring Extension

The extension of a spring is governed by Hooke's Law, which states that the force exerted by a spring is directly proportional to its extension, as long as the elastic limit is not exceeded. Mathematically, this is expressed as:

F = k * x

Where:

• F is the force applied to the spring (in Newtons).
• k is the spring constant (in N/m), which measures the stiffness of the spring.
• x is the extension of the spring (in meters).

Initial Setup

In your scenario, a single spring extends by 9 cm (or 0.09 m) when a load is applied. This means that:

F = k * 0.09

Cutting the Spring

When the spring is cut into three equal parts, each part will have a new spring constant. The spring constant of a spring is inversely proportional to its length. Therefore, if the original spring length is divided by three, the new spring constant for each part becomes:

k' = 3k

This means each of the three new springs is stiffer than the original spring.

Connecting Springs in Parallel

When springs are connected in parallel, the effective spring constant of the system is the sum of the individual spring constants:

K_eff = k' + k' + k' = 3k'

Substituting for k', we get:

K_eff = 3(3k) = 9k

Calculating the New Extension

Now, we can use Hooke's Law again to find the new extension when the same load is applied to the parallel arrangement:

F = K_eff * x'

Substituting the effective spring constant:

F = 9k * x'

Since we know from the original setup that F = k * 0.09, we can set the two equations equal to each other:

k * 0.09 = 9k * x'

Dividing both sides by k (assuming k is not zero), we have:

0.09 = 9 * x'

Solving for x', we find:

x' = 0.09 / 9 = 0.01 m

Final Result

Thus, the extension of the spring system when the three parts are connected in parallel under the same load is:

1 cm or 10 mm.

This illustrates how cutting a spring and rearranging it can significantly alter its behavior under load, showcasing the principles of elasticity and spring mechanics effectively.