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Answered on 18 Apr Learn Work and energy
Nazia Khanum
Derivation of the Formula for Potential Energy
Introduction to Potential Energy: Potential energy is the energy possessed by an object due to its position relative to other objects. It is a fundamental concept in physics and is crucial in understanding various phenomena, including gravitational and elastic potential energy.
Gravitational Potential Energy: Gravitational potential energy (PEg)(PEg) is the energy stored in an object due to its position in a gravitational field. The formula for gravitational potential energy is derived based on the gravitational force between two objects.
Formula for Gravitational Potential Energy: The formula for gravitational potential energy is given by:
PEg=mghPEg=mgh
Where:
Derivation of Gravitational Potential Energy Formula: The derivation starts with the definition of work done (WW) against gravity to raise an object to a height hh:
W=F⋅dW=F⋅d
Where:
In the case of lifting an object against gravity, the force required is the gravitational force (F=mgF=mg), and the displacement is the vertical height hh.
W=mg⋅hW=mg⋅h
Since work done is equal to the change in potential energy, we can equate it to the change in gravitational potential energy (ΔPEgΔPEg):
ΔPEg=WΔPEg=W
ΔPEg=mghΔPEg=mgh
This is the formula for gravitational potential energy, denoted by PEgPEg.
Conclusion: The formula for potential energy, particularly gravitational potential energy, is derived from the work done against gravity to raise an object to a certain height. It is given by PEg=mghPEg=mgh, where mm is the mass of the object, gg is the acceleration due to gravity, and hh is the height or distance from a reference point. Understanding this formula is essential in various fields of physics and engineering.
Answered on 18 Apr Learn Work and energy
Nazia Khanum
Definition of Power
Power is defined as the rate at which work is done or energy is transferred or converted. It measures how quickly energy is transferred or converted from one form to another.
Unit of Power
The unit of power is the watt (W), named after the Scottish engineer James Watt.
Other units of power include:
Answered on 18 Apr Learn Work and energy
Nazia Khanum
Understanding Potential Energy
Potential energy is a fundamental concept in physics that refers to the energy possessed by an object due to its position or configuration relative to other objects. It's essentially the energy stored within a system that can be converted into other forms of energy. Understanding potential energy is crucial in various fields, including physics, engineering, and chemistry.
Types of Potential Energy
Potential energy manifests in different forms depending on the nature of the system and the forces involved. Here are some common types of potential energy:
Gravitational Potential Energy
Elastic Potential Energy
Chemical Potential Energy
Electrostatic Potential Energy
Nuclear Potential Energy
Understanding the various forms of potential energy is essential for analyzing physical systems, predicting behaviors, and engineering applications across different domains.
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Answered on 18 Apr Learn Work and energy
Nazia Khanum
Work and Energy Relationship
Introduction Work and energy are fundamental concepts in physics that are closely related to each other. Understanding their relationship is crucial in comprehending various physical phenomena.
Definition of Work Work, in the context of physics, is defined as the product of force applied on an object and the displacement of the object in the direction of the force. Mathematically, it is represented as:
Work=Force×Displacement×cos(θ)Work=Force×Displacement×cos(θ)
where:
Definition of Energy Energy is the capacity to do work. It exists in various forms such as kinetic energy, potential energy, thermal energy, etc. The total energy of a system remains constant in an isolated system according to the law of conservation of energy.
Relationship Between Work and Energy The relationship between work and energy can be understood through the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Mathematically, it can be expressed as:
Work=ΔKinetic EnergyWork=ΔKinetic Energy
This theorem implies that when work is done on an object, it either gains or loses kinetic energy depending on the direction of the force applied.
Forms of Energy Energy exists in various forms, including:
Conservation of Energy According to the law of conservation of energy, energy can neither be created nor destroyed, it can only be converted from one form to another. This principle is crucial in understanding various physical phenomena and is a fundamental concept in physics.
Applications The relationship between work and energy finds applications in various fields, including:
Conclusion Work and energy are interconnected concepts in physics. The work-energy theorem provides a fundamental relationship between the work done on an object and the change in its kinetic energy. Understanding this relationship is essential for analyzing and predicting the behavior of physical systems.
Answered on 18 Apr Learn Work and energy
Nazia Khanum
Example of Negative Work Done on an Object
Introduction: In physics, work done on an object is defined as the energy transferred to or from the object by means of a force acting on it. When the force and the displacement are in the same direction, work done is considered positive, whereas when they are in opposite directions, work done is negative.
Example: Lifting an Object Upwards:
Example: Lowering an Object Downwards:
Conclusion: In conclusion, when an object is moved in a direction opposite to the force applied, the work done on the object is negative. This concept is crucial in understanding the transfer of energy and the behavior of objects under the influence of forces.
Answered on 18 Apr Learn Motion
Nazia Khanum
i) Motion of a Car around a Curve:
ii) Motion of an Electron Orbiting around a Nucleus:
Answer to Question (b):
Given:
To Calculate:
Solution:
Convert the time from hours to seconds since speed is measured in meters per second.
Apply the formula for the speed of an object in circular motion:
Substitute the given values into the formula:
Result:
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Answered on 18 Apr Learn Motion
Nazia Khanum
Understanding Uniform and Non-Uniform Motion
Uniform Motion: Uniform motion refers to the movement of an object at a constant speed in a straight line, maintaining the same velocity throughout its journey. In this type of motion, the object covers equal distances in equal intervals of time.
Example of Uniform Motion:
Non-Uniform Motion: Non-uniform motion occurs when an object changes its speed or direction over time. Unlike uniform motion, the velocity of an object in non-uniform motion is not constant; it may vary at different points during its journey.
Example of Non-Uniform Motion:
Answered on 18 Apr Learn Motion
Nazia Khanum
Answered on 18 Apr Learn Motion
Nazia Khanum
Uniform Acceleration
Definition: Uniform acceleration refers to a situation where an object's velocity changes at a constant rate over time. In other words, the object's speed increases or decreases by the same amount in each successive equal interval of time.
Acceleration of a Body with Uniform Velocity: When a body is moving with uniform velocity, its acceleration is zero. This means that the object maintains a constant speed and direction, hence no change in velocity, and consequently, no acceleration.
Magnitude of Displacement for a Particle Moving Over Three Quarters of a Circle
Given:
Calculation:
Circumference of the Circle:
Three Quarters of the Circle:
Magnitude of Displacement:
Result:
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Answered on 18 Apr Learn Motion
Nazia Khanum
Solution:
Given Data:
Conversion: To perform calculations, we need to convert velocities from km/h to m/s.
Conversion: 1 km/h = 13.63.61 m/s
Converting Initial Velocity: u=54 km/h×13.6=15 m/su=54km/h×3.61=15m/s
Converting Final Velocity: v=72 km/h×13.6=20 m/sv=72km/h×3.61=20m/s
Formula: a=v−uta=tv−u
Substituting Values: a=20 m/s−15 m/s10 sa=10s20m/s−15m/s
Calculation: a=5 m/s10 s=0.5 m/s2a=10s5m/s=0.5m/s2
Formula: s=ut+12at2s=ut+21at2
Substituting Values: s=(15 m/s×10 s)+12×0.5 m/s2×(10 s)2s=(15m/s×10s)+21×0.5m/s2×(10s)2
Calculation: s=(150 m)+0.5×5×100=150+250=400 ms=(150m)+0.5×5×100=150+250=400m
Answer: (i) Acceleration a=0.5 m/s2a=0.5m/s2 (ii) Distance Covered s=400 ms=400m
Therefore, the bus accelerates at 0.5 m/s20.5m/s2 and covers a distance of 400 m400m during this interval.
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