Two wires are made of the same material and have the same volume. The first wire has cross-sectional area $A$ and the second wire has cross-sectional area $3A$. If the length of the first wire is increased by $\Delta l$ on applying a force $F$, how much force is needed to stretch the second wire by the same amount ?

Question

A 2 $\text{kg}$ mass is attached to a spring with spring constant $ k = 200, \text{N/m} $. If the mass is displaced by $ 0.1, \text{m} $, what is the potential energy stored in the spring?

Answer 1

The problem involves two wires made of the same material with equal volumes but different cross-sectional areas. We need to determine the force required to stretch the second wire by the same amount as the first wire when a known force is applied.

To solve this, we will use the concept of Young's modulus, which is defined as:

Y = \frac{F \cdot L}{A \cdot \Delta l}

Where:

F is the force applied,
L is the original length of the wire,
A is the cross-sectional area,
\Delta l is the change in length (extension),
Y is Young's modulus (a constant for a given material).

Since the volumes of the wires are the same:

V = A_1 \cdot L_1 = A_2 \cdot L_2

Given:

For the first wire: A_1 = A and length L_1
For the second wire: A_2 = 3A and length L_2

Since the volumes are equal:

A \cdot L_1 = 3A \cdot L_2 \implies L_2 = \frac{L_1}{3}

The extension \Delta l is the same for both wires. Using Young's modulus, the force required for the first wire is:

F = \frac{Y \cdot A \cdot \Delta l}{L_1}

For the second wire, let's denote the force as F_2:

F_2 = \frac{Y \cdot (3A) \cdot \Delta l}{L_2}

Substituting L_2 = \frac{L_1}{3}:

F_2 = \frac{Y \cdot 3A \cdot \Delta l}{L_1 / 3} = \frac{9 \cdot Y \cdot A \cdot \Delta l}{L_1}

Thus, F_2 = 9 \cdot F.

Conclusion: The force required to stretch the second wire by the same amount is 9 F. Therefore, the correct answer is 9 F.

Answer 2