Physics JAMB

Elasticity

Overzicht

Welcome to the course material on Elasticity in Physics. This topic delves into the fascinating world of materials and their response to external forces. Understanding elasticity is crucial as it helps us comprehend how materials deform and return to their original shape when forces are applied and removed.

One of the key aspects covered in this topic is the force-extension curve, which provides valuable insights into a material's behavior under stress. This curve typically illustrates the relationship between applied force and resulting extension, showcasing important points such as the elastic limit, yield point, and breaking point. These critical points help us determine the maximum stress a material can endure before permanent deformation occurs.

Hooke's Law is another fundamental concept within elasticity that states the extension of a material is directly proportional to the applied force, as long as the elastic limit is not surpassed. This law is pivotal in understanding how materials behave within their linear elastic range and is often expressed as F = kx, where F is the force applied, x is the extension, and k is the material's stiffness constant.

Furthermore, Young's Modulus is a crucial parameter for materials, representing their stiffness and ability to withstand deformation. It quantifies the ratio of stress to strain in a material and is a key characteristic used to compare the elasticity of different substances.

Practical measurements of force are often carried out using a spring balance, a device specifically designed for measuring forces through the extension of a spring. By utilizing the principles of elasticity, spring balances provide accurate force measurements, making them indispensable tools in physics laboratories.

When studying springs and elastic strings, it is essential to calculate the work done per unit volume in these elements. Work done in such structures plays a significant role in understanding energy transfer and deformation processes, providing valuable insights into the behavior of elastic materials.

In conclusion, the topic of Elasticity offers a profound understanding of how materials respond to external forces, highlighting key concepts such as force-extension curves, Hooke's Law, Young's Modulus, and practical force measurement techniques using spring balances. By mastering these concepts, we can explore the intricate world of material science and its implications in various fields of physics and engineering.

Doelstellingen

  1. Use Spring Balance to Measure Force
  2. Interpret Force-Extension Curves
  3. Interpret Hooke’s Law and Young’s Modulus of a Material
  4. Determine the Work Done in Spring and Elastic Strings

Lesnotitie

Elasticity is a fundamental concept in physics that describes the ability of a material to return to its original shape and size after the removal of a force that causes deformation. Materials that exhibit high elasticity can stretch or compress significantly and then return to their original shape without permanent deformation. This ability is crucial in a wide range of applications, from everyday objects like rubber bands to sophisticated engineering materials like those used in building bridges.

Lesevaluatie

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  1. What is the term used to describe the point beyond which a material will not return to its original shape once the deforming force is removed? A. Elastic limit B. Yield point C. Breaking point D. Young's modulus Answer: C. Breaking point
  2. Which law states that the extension of a material is directly proportional to the applied force within the elastic limit? A. Hooke's law B. Newton's law C. Boyle's law D. Ohm's law Answer: A. Hooke's law
  3. Which device is commonly used to measure force by utilizing the extension or compression of a spring? A. Barometer B. Hydrometer C. Spring balance D. Thermometer Answer: C. Spring balance
  4. What is the physical quantity that measures the amount of energy transferred to a material when work is done on it per unit volume? A. Pressure B. Density C. Young's modulus D. Strain energy density Answer: D. Strain energy density
  5. Which physical quantity describes the ratio of stress to strain in a material and indicates its stiffness? A. Elastic limit B. Hooke's constant C. Young's modulus D. Strain energy Answer: C. Young's modulus

Aanbevolen Boeken

Physics for Scientists and Engineers
Physics for Scientists and Engineers
Ondertitel: Mechanics
Uitgever: Pearson
Jaar: 2017
ISBN: 978-0136273048
Fundamentals of Physics
Fundamentals of Physics
Ondertitel: Volume 1
Uitgever: Wiley
Jaar: 2020
ISBN: 978-1119655958

Herhalingsvragen

Benieuwd hoe eerdere vragen over dit onderwerp eruitzien? Hier zijn een aantal vragen over Elasticity van voorgaande jaren.

JAMB UTME - Physics - 2024

Vraag 1 Verslag

A force of 10N extends a spring of natural length 1m by 0.02m, calculate the length of the spring when the applied force is 40N.
Antwoorddetails

To solve this problem, we will use Hooke's Law. Hooke's Law states that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, it is represented as:


F = k * x


where:

  • F is the force applied to the spring.
  • k is the spring constant.
  • x is the extension or compression of the spring from its natural length.

Firstly, we need to find the spring constant k. We know that a force of 10N extends the spring by 0.02m. Therefore, using Hooke's Law:


10N = k * 0.02m


From this, we can solve for k:


k = 10N / 0.02m = 500N/m


Now that we have determined the spring constant, let's calculate the extension caused by a force of 40N:


Using Hooke's Law again:


F = k * x

40N = 500N/m * x


Solving for x:


x = 40N / 500N/m = 0.08m


This means that the spring is extended by 0.08m when a force of 40N is applied. Therefore, the length of the spring (natural length plus extension) becomes:


1.00m + 0.08m = 1.08m


Thus, the **length** of the spring when the applied force is 40N is 1.08m.

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WAEC SSCE - Physics - 2021

Vraag 1 Verslag

(a)(i) State Hooke's law. (ii) A spring has a length of 0.20 m when a mass of 0.30 kg hangs on it, and a length of 0.75 nm when a mass of 1.95 kg hangs on it. Calculate the: (i) force constant of the spring; (ii) length of the spring when it is unloaded. [g = 10m/s\(^2\)]

(b)(i) What is diffusion? (ii) State two factors that affect the rate of diffusion of a substance. (iii) State the exact relationship between the rate of diffusion of a gas and its density.
(c) A satellite of mass, m orbits the earth of mass. M with a velocity, v at a distance R from the centre of the earth. Derive the relationship between the period T, of orbit and R.

 

 

Antwoorddetails

(a)(i) Hooke's law. Provided the elastic limit is not exceeded, the extension of an elastic material is directly proportional to the force (load) producing it. \(F = k e\).

(a)(ii) Spring calculation. (The second length is 0.75 m.)

Force when 0.30 kg hangs: \(F_1 = 0.30\times10 = 3\ \text{N}\), length \(L_1 = 0.20\ \text{m}\).

Force when 1.95 kg hangs: \(F_2 = 1.95\times10 = 19.5\ \text{N}\), length \(L_2 = 0.75\ \text{m}\).

Force constant:

\[ k = \frac{F_2-F_1}{L_2-L_1} = \frac{19.5-3}{0.75-0.20} = \frac{16.5}{0.55} = 30\ \text{N/m} \]

Unloaded (natural) length \(L_0\): using \(F_1 = k(L_1-L_0)\),

\[ 3 = 30(0.20 - L_0) \;\Rightarrow\; 0.20 - L_0 = 0.1 \;\Rightarrow\; L_0 = 0.10\ \text{m} \]

Force constant \(= 30\ \text{N/m}\); natural length \(= 0.10\ \text{m}\).

(b)(i) Diffusion. Diffusion is the net movement of particles (molecules or ions) of a substance from a region of higher concentration to a region of lower concentration until they are evenly spread.

(b)(ii) Two factors affecting rate of diffusion. Temperature (higher temperature gives faster diffusion); the density or molar mass of the substance (lighter/less dense substances diffuse faster). (Also the concentration gradient.)

(b)(iii) Relationship with density. The rate of diffusion of a gas is inversely proportional to the square root of its density (Graham's law):

\[ \text{rate} \propto \frac{1}{\sqrt{\rho}} \]

(c) Period-radius relationship for a satellite. The gravitational pull provides the centripetal force:

\[ \frac{GMm}{R^{2}} = \frac{mv^{2}}{R} \;\Rightarrow\; v^{2} = \frac{GM}{R} \]

The satellite covers the circumference \(2\pi R\) in one period, so \(v = \dfrac{2\pi R}{T}\). Substituting:

\[ \left(\frac{2\pi R}{T}\right)^{2} = \frac{GM}{R} \;\Rightarrow\; \frac{4\pi^{2}R^{2}}{T^{2}} = \frac{GM}{R} \] \[ \boxed{\,T^{2} = \frac{4\pi^{2}}{GM}\,R^{3}\,} \]

Hence \(T^{2} \propto R^{3}\) (Kepler's third law).

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NECO SSCE - Physics - 2009

Vraag 1 Verslag

The work done in extending a spring by 40 mm is 1.52J. Calculate the elastic constant of the spring.

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