Work 

In physics, work is the action of moving an object by a given distance with a force, which causes an energy transfer as the body is displaced in the direction of the force.

1.0Insight into Work

As mentioned above, work is the action of moving an object by a given distance with a force, which causes an energy transfer as the body is displaced in the direction of the force. The work W can be mathematically represented as the product of the force F applied and displacement d of the object in the direction of the applied force. 

$W=F\cdot d\cdot \cos \theta$

Here:

• W is the work done (in joules, J),
• F is the applied force (in newtons, N),
• d is the displacement (in meters, m),
• θ is the angle between the force and displacement vectors.

Key Concepts

Unit of Work: 

• The SI unit of work is the joule.
• 1 joule is the amount of work done when a force of 1 Newton is applied to move an object 1 meter in the direction of the force.

1J = 1Nm 

Newton(N) is the unit of force, and the meter is the displacement unit. 

Work Done by a Constant Force

When the applied force is constant on any object and acts in the direction of displacement, the work done on the object is: 

$W=F\cdot d$

If the force is applied at an angle θ, then only the component of the force in the direction of displacement does work. The formula becomes: 

$W=F\cdot d\cdot \cos \theta$

Work Done by a Variable Force

When the applied force on any object keeps on changing with displacement, the work done is calculated using the given formula: 

$W={\int }_{x_1}^{x_2}F(x)dx$

Here, x₁ and x₂ are the magnitudes of changing force. 

Dimensions of Work

The Dimensions of work are derived from the above formula of work W = F*d. In the formula, the force F has the dimensions of [MLT^-2], and the displacement d has the dimensional formula of [L]; hence the dimensions of work are: 

$W=\left[M{L}^2{T}^{-2}\right]$

Work Energy Theorem 

Work Energy theorem states that work done on an object is converted into its change in kinetic energy. In simple words, when you apply force on an objec,t it causes the object to move or speed up or slow down if moving, resulting in a change in its kinetic energy, in short, you are transferring your energy to the object in the form of kinetic energy. Mathematically, the theorem is expressed as: 

$W_{\text{net}}=\Delta K$

Here:

• W_net the net work done on the object.
• ΔK is the change in the object's kinetic energy, which is given by:

$\Delta K=K_f-K_i$

where K_f is the final kinetic energy and K_i is the initial kinetic energy of the object. 

Applications of Work-Energy Theorem

1. Lifting objects: While lifting an object, we do work against the force of gravity. The work done here is converted into potential energy (Energy due to position). Here the work energy theorem helps in calculating how much energy is needed to raise the object at a certain height. 

2. Roller Coasters: As the coaster climbs at a height, it gains potential energy, which is later converted into kinetic energy when it descends. The work energy here at play explains energy conversion between different forms. 

3. Vehicle braking: When a vehicle, say, for example, a car comes to a stop, the brakes do work against the momentum of the car to reduce its speed. Here, the kinetic energy of the car is converted into heat energy by brake pads. 

4. Wind turbines: The wind turbines work with the help of the kinetic energy of air. The wind applies a force to move the turbine blades. Hence producing mechanical energy which is then converted into electrical energy. The work-energy theorem helps in understanding the energy conversion of these turbines. 

2.0Key Differences

Difference Between Work & Energy

• Work and energy are related but not the same concepts. Work is the transfer of energy by the application of force over a distance, whereas energy is the ability to do work.
• Work is defined as the process of energy transfer, and energy is described in many forms– kinetic, potential, thermal, etc.
• Energy is a property of an object, whereas work is the action or process that changes the energy of an object.
• Work and Energy are related to each other with the Work-Energy Theorem, which states that the work done on any object is equal to the total change in its kinetic energy: 

$W=\Delta KE=K{E}_{final}-K{E}_{initial}$

Difference Between Work & Power

• Work is the energy transferred by a force that causes displacement and depends on the force applied and the displacement of the object.
• Power is the rate of doing work or transferring energy. It is the quantity of work done per unit of time.
• In mathematical terms, the relationship between work and power can be represented as: 

$P=\frac{W}{t}$

Here, P = power, W = Work, t = time in seconds(s)

• The unit of power is watt (represented by W), while the unit of work is joule (represented by J). Power measures how quickly work is done, and work, on the other hand, measures the total amount of energy transferred.

3.0Work - Solved Questions

Problem 1: A car of mass 1500 kg accelerates from rest to a speed of 20 m/s. Calculate the work done on the car by the net force during this acceleration.

Solution:

Using the Work-Energy theorem, which says: 

$W=\Delta KE=K{E}_{final}-K{E}_{initial}$

As the car starts from the rest, then KE_initial = 0. 

$W=K{E}_{final}=\frac{1}{2}m{v}^2$

$W=\frac{1}{2}\times 1500\times 20^2$

$W=750\times 400=300000J$

Problem 2: A person pushes a box with a force of 20 N at an angle of 30 to the horizontal over a distance of 10 meters. Calculate the work done.

Solution:

By using the formula: 

$W=F\cdot d\cdot \cos \theta$

$W=20\cdot 10\cdot \cos 30$

$W=20\cdot 10\cdot \frac{\sqrt{3}}{2}$

$W\approx 200\times 0.866=173.2J$

Problem 3: A force F(x)=5x² N acts on an object along the x-axis, where x is in meters. Calculate the total work done by this force as the object moves from x=0m to x=3m.

Solution:

Given that F(x) = 5x² N , x₁ = 0, and x₂ = 3m. To calculate the variable force: 

$W={\int }_{x_1}^{x_2}F(x)dx$

$W={\int }_0^{3}5x^{2}dx$

$W=5{\left[\frac{x^3}{3}\right]}_0^3$

$W=5\left(\frac{3^3}{3}-\frac{0^3}{3}\right)=5\left(\frac{27}{3}\right)$

$W=5\times 9=45J$