GSO002 Problem 3

Let the extension of the spring be $x$ and the spring constant be $k$ . The elastic force magnitude $F$ is given by Hooke's law:

F = kx

When the hanging object is in equilibrium, the gravitational force $mg$ and the elastic force $F$ balance:

mg = kx

With the total length $L$ and natural length $L_0$ , the extension is $x = L - L_0$ , so:

mg = k(L - L_0) \quad \dots \text{①}

From the data with weight $A$ (mass $m_1$ ), the extension $x_1$ is:

x_1 = L_1 - L_0 = 0.152\text{ m} - 0.112\text{ m} = 0.040\text{ m}

From equation ①:

m_1 g = k \times 0.040

Instead of finding $g$ and $k$ separately, we use the fact that extension is directly proportional to mass.

Let the extension with container $B$ (mass $m_2$ ) be $x_2$ . From the equilibrium condition:

m_2 g = k x_2

Since $g$ and $k$ are the same in both cases, the extension is simply proportional to mass:

\frac{x_2}{x_1} = \frac{m_2}{m_1}

Substituting numerical values:

x_2 = x_1 \times \frac{m_2}{m_1} = 0.040\text{ m} \times \frac{0.635\text{ kg}}{0.200\text{ kg}}

x_2 = 0.040 \times 3.175 = 0.127\text{ m}

The total length $L_2$ is the natural length plus the extension:

L_2 = L_0 + x_2 = 0.112\text{ m} + 0.127\text{ m} = 0.239\text{ m}

The required answer is $L_2$ multiplied by $1000$ :

0.239 \times 1000 = 239

Answer: 239

Precision Spring Scale for Pharmacy Use