# When a weight is

Let the different position of spring 0, 1 and 2.

Position 0 shows the situation when no load is applied on the spring

Position 1 shows the situation when a load of w1 is applied on the spring.

We can see that there is an extension of x cm from the initial position.

Position 2 shows the situation when a load of w2 is applied on the spring.

We can see that there is extension of x2 cm from the initial position.

It is given in the question that; the extension is proportional to the applied load.

So, any extension x will be proportional to the weight w that produces that extension.

We can write: x = a constant k × w

x = kw

The constant ‘k’ is called the spring constant. Every spring will have a particular value of spring constant. We want to find this constant for our spring.

For that, we adopt the following procedure:

Put a known weight w1. Measure the extension x1.

Then

Put another known weight w2. Measure the extension x2.

Then

Since k is a constant, we will get the same value for k in both the weight.

Repeat the trial with several known weights. In all cases we will get the same k.

Once k is obtained, we can make the markings on the spring balance.

The spring is fixed inside an outer casting. Markings are made on this casing.

When the spring is at Zero load position, mark that position on the casing as 0kg.

Now we want to mark the 1kg position

Let the extension for a load of 1kg be x1 kg

x1 = k × 1 kg = k

we have already calculated k. so we get x1 kg.

Measure this x1 kg from the 0kg mark on the casing and mark it as 1kg.

Now we want to mark the 2kg position

Let the extension for a load of 2kg be x2 kg

X2 = k × 2 kg = 2k

we have already calculated k. so we get x2 kg.

Measure this x2 kg from the 0kg mark on the casing and mark it as 2kg.

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