What is Work in Physics?

Introduction:

In Physics, the concept of What is Work ?plays a pivotal role in understanding how forces cause movement and how energy is transferred or transformed in physical systems. The term "work" in physics has a precise definition, different from its everyday use, where it generally refers to any physical or mental effort. In physics, work is a fundamental concept that connects force, displacement, and energy, serving as a cornerstone in mechanics, thermodynamics, and various other domains of physics.

In his article I will try to delve into the definition of What is Work in Physics?, the mathematical formulation, the different types of work, the relationship between work energy, power and various applications of the concept in understanding natural phenomena.

Explanation:

Let, us start with

What is Work in Physics?

Definition of Work in Physics

Work, in the context of physics, is defined as the process of energy transfer when a force is applied to an object, causing it to move in the direction of the force. Mathematically, work (W) is expressed as:

W=F⁻⁻>d⁻⁻>=Fd cos⁡θ

Where:

W is the work done by the force,

F⁻⁻>is the vector representing the force applied,

d⁻⁻> is the vector representing the displacement of the object,

F is the magnitude of the force,

d is the magnitude of the displacement,

θ is the angle between the force vector and the displacement vector.

Work is a scalar quantity, meaning it has magnitude but no direction. The unit of work in the International System of Units (SI) is the joule (J), where one joule is equivalent to one newton-meter (1 J=1 N⋅m)

Again, in simple words, work can be defined as, force x displacement or

W = F x d

What is 1J?

Answer: When 1N force is applied on a body which at rest and it displaced 1m in the direction of force i.e; if it achieved 1 m displaced then the work which is done is called 1J.

Some problems related on work:

Question(1): 5N force is applied on a body and it is displaced 2m in the direction of force,then find the work?

Answer: Here,

Force (F) = 5N

Displacement (d) = 2 m

∴ Work = F x d

= ( 5 x 2)

= 10 J

Question(2): How much work has to be done to increase the velocity of a car from 30 ms⁻¹ to 60 ms ⁻¹, if the mass of the car is 1500kg.

Answer: As we know that the capacity of doing work is called energy. So energy and work are same.

Here,

mass of the car (m) = 1500 kg

1st case, velocity (v₁) = 30 ms⁻¹

2nd case, velocity(v₂) = 60 ms⁻¹

∴ Required work done will be

= ½ m (v₂² - v₁²)

= ½ x 1500 ( 60² - 30² ) [putting values]

= 750 x (3600 - 900)

= 750 x 2700

= 2025000 J

= 2025 kJ [ ∵ 1000 J = 1 kJ]

Historical Context and Development

The concept of work has evolved over centuries, deeply intertwined with the development of classical mechanics. The idea of work was formalized during the 18th and 19th centuries as scientists began to understand the relationship between force, motion, and energy.

The term "work" in the scientific context was first introduced by the French mathematician Gaspard-Gustave Coriolis in the early 19th century. Coriolis was instrumental in defining work as the product of force and distance moved by the force in the direction of its action. This formalization was crucial in linking the concept of work to energy, setting the stage for the development of the law of conservation of energy.

Mathematical Formulation of Work

The mathematical expression for work, W=Fdcos⁡θ, encapsulates several critical aspects of the physical process:

a. Magnitude of Force and Displacement

The work done depends on the magnitude of the force applied and the displacement of the object. If either the force or the displacement is zero, no work is done. This means that even if a significant force is applied, if there is no movement, the work done is zero.

b. Directionality and the Cosine Component

The angle θ between the force and the displacement is crucial. The cosine component cos⁡θ determines the effective part of the force that contributes to the work done. The scenarios include:

When θ=0°(force and displacement are in the same direction), cos⁡0∘=1, and work is maximized as W=Fd.

When θ=90°(force and displacement are perpendicular), cos⁡90°=0, and no work is done as W=0.

When θ=180°(force and displacement are in opposite directions), cos⁡180°= - 1, and work is negative, indicating that the force is acting against the displacement.

Positive, Negative, and Zero Work

Positive Work: Positive Work occurs when the force component is in the same direction as the displacement, resulting in the energy being transferred to the object.

Negative Work: Negative Work Occurs when the force component is in the opposite direction to the displacement, indicating that energy is taken from the object.

Zero Work: Zero Work occurs when the force is perpendicular to the displacement or when there is no displacement at all.

Types of work in Physics:

Work can be classified into different types based on the context and the nature of the forces involved:

a. Mechanical Work

Mechanical work refers to the work done by forces that cause the movement of objects. It is the most common type of work encountered in classical mechanics and includes scenarios like lifting an object, moving a block along a surface, or pushing a car.

b. Gravitational Work

Gravitational work involves the work done by gravitational forces. For example, when an object is lifted against the gravitational pull of the Earth, work is done by the lifting force, and the object gains gravitational potential energy.

W=mgh

Where:

m is the mass of the object,

g is the acceleration due to gravity,

h is the height the object is lifted.

c. Elastic Work

Elastic work is the work done when deforming an elastic object, such as stretching or compressing a spring. Hooke's Law governs the relationship between the force applied and the deformation in such cases. The work done in stretching or compressing a spring is given by:

W = ½ kx²

Where,

k is the spring constant,

x is the displacement from the equilibrium position.

d. Work Done by a Variable Force

In many real-world scenarios, the force applied to an object may not be constant. In such cases, the work done is calculated using calculus, integrating the force over the displacement:

W= x₂

∫ F(x).dx

x₁

This approach is crucial for analyzing systems where the force changes with position, such as in gravitational fields, electric fields, or when dealing with non-linear springs.

Work-Energy Theorem

The work-energy theorem is a fundamental principle that relates the work done on an object to its kinetic energy. It states that the net work done on an object is equal to the change in its kinetic energy:

Wₙ = Δ K = ½ mvᵢ² - ½ mvₘ²

Where:Wₙis the net work done on the object,

Δ Kis the change in kinetic energy,

m is the mass of the object,

vᵢ is the initial velocity,

vₘis the final velocity.

This theorem is a powerful tool in mechanics, allowing the prediction of an object's motion and speed after a force is applied.

Work and Energy Conservation

The concept of work is deeply linked to the law of conservation of energy, one of the most fundamental principles in physics. According to this law, energy cannot be created or destroyed, only transferred or transformed from one form to another.

a. Work and Kinetic Energy

As mentioned in the work-energy theorem, the work done on an object results in a change in its kinetic energy. If positive work is done on the object, its kinetic energy increases. Conversely, if negative work is done, its kinetic energy decreases.

Some problems on Kinetic Energy.

Question(1): A body of mass 13 kg moving with an uniform velocity of 4 ms⁻¹,find its kinetic energy.

Answer: Here,

Mass of the body (m) = 13 kg

Uniform velocity (v) = 4 ms⁻¹

∴ Kinetic energy (Ke) = ½ mv²

= ½ x 13 x 4 x 4 [putting values]

= 13 x 2 x 4

= 104 J

Question(2):A car with mass 1500kg moving with a speed 60 kmh⁻¹,to stop this car what amount of work has to be done.

Answer: Here,

mass of the car (m) = 1500 kg

speed of the car (v) = 60 kmh⁻¹

= (60 x 1000) / 3600 [∵1km =1000m and 1hr = 3600s]

= 1000 / 60

= 16.666….

= 16.67 ms⁻¹ (approximately)

∴ Work has to be done to stop the car

= ½ mv²

= ½ x 1500 x 16.67 x 16.67 [putting values]

= 750 x 277.8889

= 208416.675

= 208417 J (approx.)

b. Work and Potential Energy

Work done against conservative forces like gravity or spring force is stored as potential energy. For example, lifting an object increases its gravitational potential energy, while compressing a spring increases its elastic potential energy.

Some problems related on Potential Energy:

Question(1): A ball has a mass 5kg is lifted at a height 10 meter from the ground find it’s potential energy.(Take g = 10 m/s²)

Answer: Mass (m) = 5 kg

Height (h) = 10m

g = 10 m/s²

∴ Potential Energy = mgh

= 5 x 10 x 10

= 500 J

Question(2): To lift a mass of 50 kg at some height 500 J work has been done.Find the height.(Take g = 10 m/s²)

Answer: Mass (m) = 50kg

work (w) = 500 J

g = 10m/s

Let, height = h meter

∴ Height (h) = w / mg

= 500 / (50 x 10) [ Here, work = mgh]

= 500 / 500

= 1 m

Power: The Rate of Doing Work

Power is the rate at which work is done or energy is transferred. It quantifies how quickly work is performed and is given by:

P=W/t

Where:

P is the power,

W is the work done,

t is the time taken.

The SI unit of power is the watt (W), where one watt is equivalent to one joule per second (1 W=1 J/s

Power is a crucial concept in understanding how efficiently work is performed in mechanical systems, engines, and electrical circuits.

Some problems related on Power:

Question (1): A boy has a mass of 50kg climbed up 45 stairs in 9 secs,height of each stair is 15 cm,find the power of the boy.(Take g = 10m/s²)

Answer: Mass of the boy(m) = 50kg

Height (h) = (45 x 15) cm

= (675 /100)m [∵100cm =1m]

= 6.75 m

Acceleration due to gravity(g) = 10m/s²

Time(t) = 9sec

∴ Work has been done by the boy (w)= mgh

= (50 x 10 x 6.75)[putting values]

= 3375.00

= 3375 J

∴ Power of the boy = w/t

= 3375/9 Jsec⁻¹

= 375 watt.

Question(2):A body of mass 12 kg is kept at a height from the ground,if the potential energy of the body is 480 J,then find the height of the body from the ground.(Take g =10m/s²)

Answer: 1st Method,

Here,

mass of the body (m) =12kg

Acceleration due to gravity(g) = 10m/s²

Potential energy (P.E) = 480 J

∴height of the body(h) from the ground = P.E /(mxg)

= 480/(12x10)[putting values]

= 480/120

= 4 m

2nd Method,

Here,

mass of the body (m) =12kg

Acceleration due to gravity(g) = 10m/s²

Potential energy (P.E) = 480 J

Let height of the body from the ground = h meter

A/Q,

mgh = 480

=>12 x10 x h = 480

=> 120 h = 480

=> h = 480 /120

=> h = 4

∴height of the body(h) from the ground = 4m

Work in Different Physical Systems

The concept of work extends beyond simple mechanical systems and is applicable in various fields of physics, including thermodynamics, electromagnetism, and quantum mechanics.

a. Work in Thermodynamics

In thermodynamics, work is associated with the transfer of energy between a system and its surroundings. It is often related to the expansion or compression of gases. For example, in an ideal gas undergoing expansion, the work done by the gas is given by:

W= v₂

∫ P.dv

v₁

Where:

P is the pressure of the gas,

V is the volume of the gas.

This work is crucial in understanding heat engines, refrigerators, and various other thermodynamic processes.

b. Work in Electromagnetism

In electromagnetism, work is done when a charge moves in an electric field. The work done on a charge q moving through a potential difference V is given by:

W=qV

This relationship is fundamental in the study of electric circuits, capacitors, and the energy stored in electric fields.

Some problems related work in electromagnetism:

Question(1):An electric bulb of 60w is used 6hr/day, find the energy consumed by the bulb in units in a day.

Answer: Here,

Power(P) = 60w

= 60 /1000 kw [ ∵ 1000 w = 1kw]

= .06 kw

Time (t) = 6hrs

∴ Energy consumed by the bulb(w) = P x t

= .06 x 6

= . 36 kwhr

= .36 units [∵1kwhr = 1unit]

Question(2):In a house energy consumed in a month is 250 unit,how much will be this energy in joule unit?

Answer: Energy (w) = 250 unit

= 250 kwhr [∵ 1unit = 1kwhr]

= (250 x 3.6 x 10⁶ J [1kwhr = 3.6 x 10⁶ J]

= 250 x 3.6 x 1000000 J

= 900,000,000 J

c. Work in Quantum Mechanics

In quantum mechanics, the concept of work is more abstract but still essential. Work can be related to the change in the quantum state of a system or the energy associated with a particle's wavefunction. Though the classical definition of work is not directly applicable, the principles of energy transfer and conservation remain relevant.

Practical Examples and Applications of Work

The concept of work is not just theoretical; it has numerous practical applications in everyday life and technology.

a. Lifting Objects

When we lift a heavy object, you are doing work against the force of gravity. The energy we expend is transferred to the object as gravitational potential energy, which can be released if the object falls.

b. Machines and Engines

Machines like levers, pulleys, and engines are designed to do work more efficiently. For example, an internal combustion engine does work by converting chemical energy from fuel into mechanical energy, which powers a vehicle.

c. Electrical Appliances

Electric appliances like fans, refrigerators, and computers do work by converting electrical energy into mechanical work, cooling, or processing data. The power rating of these devices indicates how much work they can do in a given time.

To Read more.......