Question

Медная проволока длиной 1= 80 см и сечением 5 = 8 мм² закреплена одним концом в подвесном устройстве, а к ее другому концу прикреплен груз массой м = 400 г. Вытянутую проволоку с грузом, отклонив до высоты подвеса, отпускают. Считая проволоку невесомой, определить ее удлинение в нижней точке траектории движения груза. Модуль Юнга для меди Е= 118 гпа.​

Answer 1

Ответ:

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Объяснение:

Answer 2

Problem Analysis

We are given a copper wire with a length of 80 cm and a cross-sectional area of 8 mm². One end of the wire is fixed to a suspension device, and a mass of 400 g is attached to the other end. The wire is stretched and released, causing the mass to move in a trajectory. We need to determine the elongation of the wire at the lowest point of the trajectory. The Young's modulus for copper is given as 118 GPa.

Solution

To solve this problem, we can use Hooke's Law, which states that the elongation of a material is directly proportional to the force applied to it, as long as the material remains within its elastic limit. Mathematically, Hooke's Law can be expressed as:

F = k * ΔL

Where: - F is the force applied to the wire - k is the spring constant (also known as the stiffness or Young's modulus) - ΔL is the change in length of the wire

In this case, the force applied to the wire is the weight of the mass attached to it, which can be calculated as:

F = m * g

Where: - m is the mass of the object (400 g) - g is the acceleration due to gravity (approximately 9.8 m/s²)

Now, we can rearrange Hooke's Law to solve for ΔL:

ΔL = F / k

To find the spring constant (k), we can use the formula:

k = (E * A) / L

Where: - E is the Young's modulus for copper (118 GPa) - A is the cross-sectional area of the wire (8 mm²) - L is the original length of the wire (80 cm)

Let's calculate the elongation of the wire at the lowest point of the trajectory.

Calculation

First, let's convert the given values to SI units: - Length of the wire (L) = 80 cm = 0.8 m - Cross-sectional area (A) = 8 mm² = 8 × 10⁻⁶ m² - Mass of the object (m) = 400 g = 0.4 kg - Young's modulus for copper (E) = 118 GPa = 118 × 10⁹ Pa - Acceleration due to gravity (g) = 9.8 m/s²

Now, let's calculate the spring constant (k): k = (E * A) / L

Substituting the values: k = (118 × 10⁹ Pa * 8 × 10⁻⁶ m²) / 0.8 m

Simplifying the expression: k = 118 × 10⁶ N/m

Next, let's calculate the force applied to the wire (F): F = m * g

Substituting the values: F = 0.4 kg * 9.8 m/s²

Simplifying the expression: F = 3.92 N

Finally, let's calculate the elongation of the wire (ΔL): ΔL = F / k

Substituting the values: ΔL = 3.92 N / 118 × 10⁶ N/m

Simplifying the expression: ΔL = 3.32 × 10⁻⁵ m

Therefore, the elongation of the wire at the lowest point of the trajectory is approximately 3.32 × 10⁻⁵ m.

Answer

The elongation of the copper wire at the lowest point of the trajectory, when a mass of 400 g is attached to it, is approximately 3.32 × 10⁻⁵ m.