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A wire suspended vertically from one of its ends is stretched by attaching a weight of \(200 \mathrm{~N}\) to the lower end. The weight stretches the wire by \(1 \mathrm{~mm}\). The elastic energy stored in the wire is [2003] (A) \(0.2 \mathrm{~J}\) (B) \(10 \mathrm{~J}\) (C) \(20 \mathrm{~J}\) (D) \(0.1 \mathrm{~J}\)

Short Answer

Expert verified
The elastic potential energy stored in the wire is \(0.1 J\), which corresponds to option (D).

Step by step solution

01

Convert the elongation to meters

Since the elongation is given in millimeters, we need to convert it to meters (SI unit). \[1 mm = 0.001 m \] So, the elongation x = 0.001 m.
02

Find the spring constant k

We'll use Hooke's Law to find the spring constant k. Hooke's Law states that the force (F) required to stretch or compress a spring is proportional to its elongation (x): \[F = kx\] We already know the force (F = 200 N) and the elongation (x = 0.001 m). Now, we can solve for the spring constant k. \[k = \frac{F}{x}\]
03

Calculate the spring constant k

Plug in the values for F and x. \[k = \frac{200 N}{0.001 m} = 200,000 N/m\]
04

Calculate the elastic potential energy U

Now that we have the spring constant (k = 200,000 N/m) and the elongation (x = 0.001 m), we can find the elastic potential energy (U). Use the formula for elastic potential energy: \[U = \frac{1}{2}kx^2\] Plug in the values for k and x, \[U = \frac{1}{2}(200,000 N/m) (0.001 m)^2\]
05

Solve for the elastic potential energy U

Perform the calculations: \[U = 0.1 J\] The elastic potential energy stored in the wire is \(0.1 J\), which corresponds to option (D).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hooke's Law
When discussing the stretching of a wire or spring due to an applied force, Hooke's Law is a fundamental principle. This law describes how elastic materials respond to forces. It states that the force required to either stretch or compress a spring is directly proportional to its elongation or compression.

This can be mathematically expressed as:
  • \[ F = kx \]
where:
  • \( F \) is the force applied,
  • \( x \) is the elongation (or compression), and
  • \( k \) is the spring constant.
In simple terms, Hooke's Law implies that the force needed to change the length of a spring is linearly related to how much you stretch or compress it.

In the exercise, a weight of 200 N was attached, resulting in the wire stretching. With this constant force and considering the constant proportionality defined by Hooke's Law, we can calculate other variables, such as elongation or energy.
Spring Constant
The spring constant (\( k \)) is a measure of a spring's stiffness. A higher spring constant means the spring is stiffer, requiring more force to stretch or compress it the same amount compared to a spring with a lower \( k \).

In the exercise, we used the given values of force (200 N) and elongation (0.001 m) to calculate \( k \) using the Hooke’s Law formula:

  • \[ k = \frac{F}{x} \]
Substituting the known values:
  • \[ k = \frac{200 \, N}{0.001 \, m} = 200,000 \, N/m \]
This large spring constant indicates the wire is quite stiff, as it requires significant force to produce even a small deformation. Determining \( k \) is pivotal for calculating other quantities such as the elastic potential energy stored in the material.
Elongation Measurement
Understanding how to measure and interpret elongation is key when working with elastic materials. Elongation represents the change in length from the material's original state. In calculations or practical scenarios, it is important to use consistent units, typically meters in the International System of Units (SI).

In the given exercise, elongation was initially provided in millimeters (1 mm), which was converted to meters as:
  • \[ 1 \, mm = 0.001 \, m \]
This conversion ensures accuracy in further calculations, such as determining the spring constant or elastic potential energy. Recognizing the original and altered lengths gives vital insight into the stress and strain relationship.

Accurate elongation measurements allow for precise computations of other properties in material science and engineering contexts. It serves as the foundation for discussing how materials behave under different forces.

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Most popular questions from this chapter

A bimetallic strip is made of aluminium and steel \(\left(\alpha_{A l}>\alpha_{\text {steel }}\right)\). On heating, the strip will (A) remain straight. (B) get twisted. (C) bend with aluminium on concave side. (D) bend with steel on concave side.

An iron tyre is to be fitted onto a wooden wheel \(1.0 \mathrm{~m}\) in diameter. The diameter of the tyre is \(6 \mathrm{~mm}\) smaller than that of wheel. The tyre should be heated so that its temperature increases by a minimum of (coefficient of volumetric expansion of iron is \(3.6 \times 10^{-5} /{ }^{\circ} \mathrm{C}\) ) (A) \(167^{\circ} \mathrm{C}\) (B) \(334^{\circ} \mathrm{C}\) (C) \(500^{\circ} \mathrm{C}\) (D) \(1000^{\circ} \mathrm{C}\)

A bimetallic strip is formed by two identical strips, one of copper and the other of brass. The coefficients of linear expansion of the two metals are \(\alpha_{C}\) and \(\alpha_{B}\). On heating, the temperature of the strip goes up by \(\Delta T\) and the strip bends to form an arc of radius of curvature \(R\). Then \(R\) is (A) proportional to \(\Delta T\). (B) inversely proportional to \(\Delta T\). (C) proportional to \(\left|\alpha_{B}-\alpha_{C}\right|\). (D) inversely proportional to \(\left|\alpha_{B}-\alpha_{C}\right|\).

At what temperature, the Fahrenheit and the Celsius scales will give numerically equal (but opposite in sign) values? (A) \(-40^{\circ} \mathrm{F}\) and \(40^{\circ} \mathrm{C}\) (B) \(11.43^{\circ} \mathrm{F}\) and \(-11.43^{\circ} \mathrm{C}\) (C) \(-11.43^{\circ} \mathrm{F}\) and \(+11.43{ }^{\circ} \mathrm{C}\) (D) \(+40^{\circ} \mathrm{F}\) and \(-40^{\circ} \mathrm{C}\)

Two rods of length \(L_{1}\) and \(L_{2}\) are made of materials whose coefficients of linear expansion are \(\alpha_{1}\) and \(\alpha_{2}\). If the difference between the two lengths is independent of temperature. (A) \(\left(L_{1} / L_{2}\right)=\left(\alpha_{1} / \alpha_{2}\right)\) (B) \(\left(L_{1} / L_{2}\right)=\left(\alpha_{2} / \alpha_{1}\right)\) (C) \(L_{1}^{2} \alpha_{1}=L_{2}^{2} \alpha_{2}\) (D) \(\alpha_{1}^{2} L_{1}=\alpha_{2}^{2} L_{2}\)

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