/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 104 A wire is fixed at the upper end... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A wire is fixed at the upper end and stretches by length \(\ell\) by applying a force \(F\). The work done in stretching is(A) \(\frac{F}{2 \ell}\) (B) \(F \ell\) (C) \(2 F \ell\) (D) \(\frac{F \ell}{2}\)

Short Answer

Expert verified
The correct answer is: (B) \(F \ell\).

Step by step solution

01

Identify the formula for work done

We will use the formula for work done: \(W = F \cdot d \cdot \cos \theta\). In this case, since the force is applied in the same direction as the displacement (stretching), the angle \(\theta\) is \(0\). Step 2: Simplify the formula using the angle
02

Simplify the formula using the angle

Since \(\theta\) is \(0\), we can simplify the formula by substituting \(\cos(0)\), which is \(1\). So our simplified formula becomes \(W = F \cdot d\). Step 3: Substitute the length of stretching
03

Substitute the length of stretching

The wire stretches by a length \(\ell\), so we can substitute this value for the distance \(d\) in our simplified formula. This gives us the formula \(W = F \cdot \ell\). Step 4: Find the correct option
04

Find the correct option

Comparing our derived formula \(W = F \cdot \ell\) with the given options, we can see that this matches option (B). Therefore, the work done in stretching the wire is \(F \ell\). The correct answer is: (B) \(F \ell\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Force and Displacement
In physics, understanding the relationship between force and displacement is essential to comprehend how work is done. Force is a push or pull on an object, and displacement measures how far the object moves. When we apply a force to move an object, like stretching a wire, we effectively do work on the object. The basic formula for work, which is central in these calculations, is given by:
  • \( W = F \cdot d \cdot \cos \theta \)
Here, \( W \) is work, \( F \) is the force applied, \( d \) is the distance the object moves (or displacement), and \( \theta \) is the angle between the force and displacement.
When force is applied in the same direction as the movement, such as stretching a wire, it simplifies the situation because the displacement and force align, making the angle, \( \theta \), zero. This concept greatly simplifies calculations, as you'll see in elasticity and stretching.
Elasticity and Stretching
Elasticity refers to a material's ability to return to its original shape after being stretched or compressed. When dealing with a wire or similar elastic materials, applying force results in stretching. In terms of work, when stretching an elastic material, the length it stretches (\( \ell \)) is crucial since it directly relates to how much work is done.
  • By substituting the stretch length into our work formula, we get \( W = F \cdot \ell \).
This work depicts the energy transferred to the wire causing it to stretch. The result is visible in how much the wire elongates, impacting factors like its elasticity and the amount of energy required to achieve that change. Recognizing how elasticity affects the stretching process helps us understand mechanical stress and energy transfer more deeply.
Angle in Work Calculations
The angle \( \theta \) between the force and the direction of displacement plays a significant role in determining the work done. In the context of stretching a wire, the angle is crucial, specifically when simplifying the work formula. When the force applied aligns perfectly with the displacement, meaning they are in the same direction, the angle \( \theta = 0^\circ \). This means:
  • \( \cos(0) = 1 \)
  • This simplifies the work formula to \( W = F \cdot d \).
The perfectly aligned angle effectively eliminates the angle component from the equation, streamlining calculations significantly. This direct relationship clarifies the energy transfer when stretching occurs. Understanding this alignment helps students grasp why certain physics problems, like our wire example, become easier to calculate.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The volume of a liquid flowing per sec out of an orifice at the bottom of a tank does not depend upon (A) the height of the liquid above the orifice. (B) the acceleration due to gravity. (C) the density of the liquid. (D) the area of the orifice.

An object is floating in a liquid, kept in a container. The container is placed in a lift. Choose the correct option(s). (A) Buoyant force increases as lift accelerates up. (B) Buoyant force decreases as lift accelerates up. (C) Buoyant force remains constant as lift accelerates. (D) The fraction of solid submerged into liquid does not change.

A rectangular block of density \(\rho\), base area \(A\), and height \(h\) is kept on a spring. The lower end of spring is fixed on the bottom of an empty vessel of base area \(2 A\). The block compresses the spring by \(h / 4\) at equilibrium. The vessel is then slowly filled by a liquid of density\(2 \rho\) till the spring becomes relaxed. The block is then slowly pushed inside the liquid till it is immersed completely. Work done to push the block completely inside is \(W_{1}\), work done by gravity on the block is \(W_{2}\), and work done by upthrust is \(W_{3}\). Match the following based on the above statements. Column-I \(\quad\) Column-II (A) \(\left|W_{1}\right|\) (1) \(\frac{5 \rho g h^{2} A}{8}\) (B) \(\left|W_{2}\right|\) (2) \(\frac{\rho g h^{2} A}{8}\) (C) \(\left|W_{3}\right|\) (3) \(\frac{3 \rho g h^{2} A}{4}\) (4) \(\frac{\rho g h^{2} A}{4}\)

A structural steel rod has a radius of \(10 \mathrm{~mm}\) and a length of \(1 \mathrm{~m}\). A \(100 \mathrm{kN}\) force \(F\) stretches it along its length. The strain in rod is \(1.59 \times 10^{-n}\) then the value of \(\mathrm{n}\) is. Given that the Young's modulus \(E\) of the structural steel is \(2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{-2}\).

A wire of length \(L\) and cross-sectional area \(A\) is made of a material of Young's modulus \(Y .\) If the wire is stretched by an amount \(x\), the work done is (A) \(\frac{Y A x^{2}}{2 L}\) (B) \(\frac{Y A x}{2 L^{2}}\) (C) \(\frac{Y A x}{2 L}\) (D) \(\frac{Y A x^{2}}{L}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.