In previous lesson, we have discussed work. In this lesson, we are going to continue with energy. First of all, we will discuss the definition of energy. After that we will discuss 2 types of energy: the kinetic energy and potential energy. There are 2 types of potential energy: gravitational potential energy and elastic potential energy. After that, we will discuss the principle of conservation of energy and the work energy model.

Energy can be defined as the capacity to do work. Work is done when energy is converted from one form to another.

Energy is a scalar quantity. The unit of energy is Joule. We can see that the unit of energy is same as the unit of work done.

### Kinetic Energy

Kinetic energy is the energy of motion. When an object moves, it possesses kinetic energy. Energy is denoted by the symbol, E. We can use a symbol E sub K to represent kinetic energy, where K denotes kinetic.

In equation, kinetic energy is given by E_{K} = ½ mv^{2}, where E sub K is the kinetic energy, m is the mass of the moving object and v is the velocity or speed of the object.

For example, when a car of mass 800kg is moving at a speed of 80 m/s, the kinetic energy is ½ x 800 x (80)^{2}, which is equal to 2,560,000J. The kinetic energy possessed by the car is two millions five hundred and sixty Joules.

Take notes that if you want to use Joule as the unit of your final answer, the unit of your mass must be kg and the unit of speed must be m/s. If the speed is given in km/h, then you need to convert it into m/s.

Let’s see this example. A car of mass 500kg accelerates from 20m/s to 50m/s in 3s. Find the kinetic energy change of the car.

First of all, let’s list down all the information that we have. The mass of the car is 500kg; the initial velocity is 20 m/s; the final velocity is 50 m/s. Well, the equation for kinetic energy is E_{K} = ½ mv^{2}. One of the mistake possibly made by some students is they take the v as the velocity change.

For example, the velocity change is 50 m/s – 20 m/s. Therefore, they will write E_{K} = ½ x 500 x 30^{2}, which is equal to 225,000J.

Well, this is not correct. The kinetic energy change should be equal to final kinetic energy – initial kinetic energy, which is equal to ½ mv^{2} – ½ mu^{2}. Factorise the equation, it become ½ m(v^{2} – u^{2}). Plug in the numbers. The kinetic energy change is equal to 525,000J.

In conclusion, the kinetic energy change can be calculated by the equation ΔE_{K} = ½m(v^{2} – u^{2}). The symbol delta is used to denote change. Delta E sub K means, change in kinetic energy.

### Potential Energy

Potential energy is the energy stored in an object due to its vertical position or state. There are 2 types of potential energy: the gravitational potential energy and the elastic potential energy.

Gravitational potential energy is the energy stored in an object as the result of its vertical position. The energy is stored as the result of the gravitational attraction of the Earth for the object. For example, if we lift a mass from ground level to a height of h, some potential energy will be stored inside the mass.

Gravitational potential energy can be represented by the equation E_{P} = mgh, where m is the mass of the object, g is the gravitational acceleration or gravitational field strength and h is the height of the object from a reference level. Normally, we take the reference level at the ground level.

For example, if we lift an object of mass 2kg to a height of 1.5m, the potential energy stored inside the object will be Ep = mgh, where m = 2, g = 10 and h = 1.5. Therefore, Ep = 30J.

Well, you may find this mgh look familiar. Yes, in previous lesson, we have learned that the work done by gravity and against gravity is also given by the same equation, W = mgh. Actually, work done and energy are interrelated. We will come to that part later in this chapter.

Elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing. The equation for elastic potential energy is ½ Fx or ½ kx^{2}, where F is the force used to stretch or compress the elastic object, x is the extension or compression caused by the force, k is the spring constant. We will discuss elastic potential energy in another lesson called elasticity.

In the equation E_{P} = mgh, h is the vertical height of the object. The potential energy stored in an object is independent of the path it moves to the place. For example, if an object of mass 5kg is moved to the top of a table from the ground and the distance traveled by the object is 2m while the height of the table is 1.5m. h is the 1.5m height but not the distance 2m. How the object moved to the table top is not important. When calculating gravitational energy, we only interested in the vertical height.

Let’s see another example. A 3 kg ball moves from point A to point B and then to point C. How much is the potential energy gain of the ball at point C?

Well, the vertical height of C from A is 20 – 8, which is equal to 12m. The mass of the ball is 3kg. Potential energy, E_{P} = mgh. Plug in the numbers. The potential energy at point C is equal to 360J. The Principle of Conservation of Energy states that energy cannot be created or destroyed. It can be transformed from 1 form to another, but the total energy in the system will be constant.

For example, an object of mass 5kg is released from point A which is 20m from the ground level. The potential energy of the object will transform into kinetic energy when the object falls down freely. At point A, the potential energy E_{P} = mgh, where m = 5kg; g = 10 m/s^{2} and h = 20m. Therefore the E_{P} = 1000J. Let’s write it here E_{P} = 1000J. At this point, the velocity of the object is zero. Therefore the kinetic energy is zero. The sum of the energy is 1000J.

When the object falls to point B, which is 5m lower, the potential energy become 750J and the kinetic energy will become 250J. The sum of the energy is still 1000J.

When the object falls to point C, which is 10m lower, the potential energy become 500J and the kinetic energy will become 500J as well. Again, the sum of the energy is still 1000J.

When the object reach the ground, the potential energy become zero, while the kinetic energy become 1000J. The sum of the energy is still 1000J.

This is what we mean the sum of the energy in the system remain constant. The kinetic energy may change, the potential energy may change, but sum of the energy will remain unchanged. This is what the Principle of Conservation of Energy all about.

Let’s study a few more examples. A coconut falls from a height of 5m. What is the velocity of the coconut right before it strikes the ground?

Well, this is a free falling question. We can solve it by using the equations of uniform acceleration. Now, this time, let’s see whether we can solve it by using the principle of conservation of energy.

When the coconut is at a height h, its potential energy, E_{P} = mgh. When it falls down, the potential energy transforms into kinetic energy. The equation for kinetic energy is E_{K} = ½ mv^{2}. If we ignore the air resistance, we can assume that all the potential energy is transformed into kinetic energy at the moment right before the coconut strike the ground. Therefore, E_{K} = E_{P}. E_{K} = ½ mv^{2}; E_{P} = mgh. Cancel the m. v^{2} = 2gh. v equal to the root of 2gh. g is equal to 10 m/s^{2}; h = 5m. Plug in these numbers into the equation. Therefore, v equal to the root of 100, which is equal to 10 m/s. The velocity is 10 m/s.

The kinetic energy gain is equal to the potential energy lose. This is a typical example of using the principle of conservation of energy to find the velocity of an object.

An object of mass 5kg has initial velocity of 5 m/s. The object moves on a rough surface and comes to stop after travelling 10m. Calculate the heat energy produced during the motion.

First of all, let’s list down all the information given. The mass of the object, m = 5kg. Initial velocity = 5 m/s. Final velocity = 0 m/s because the object finally stopped after the motion. When a moving object stops due to frictional force, all its kinetic energy converted into heat energy. The amount of kinetic energy lost will be equal to the amount of heat energy produced. We write this as E_{H} = E_{K}, where E_{K} is the kinetic energy and E_{H} is the heat energy. The kinetic energy lost equal to ½ mv^{2 }– ½ mu^{2}. Plug in the numbers. Therefore, E_{H} = |-62.5J|, which is equal to 62.5J. The heat energy produced is 62.5J.

The negative sign shows that there is kinetic energy lose. Modulus is used here because we are only interested in the change in kinetic energy. Whether it is kinetic energy lose or kinetic energy gain is not important to us.

Energy is the capacity to do work. In other words, we need energy to do work. Work is done to convert one type of energy to another type of energy. The amount of work done is equal to the amount of energy being converted or energy change.

For example, when an object of mass 1kg falls from a height of 10m, 1000J of work is done to convert 1000J of potential energy into 1000J of kinetic energy. The amount of work done is 1000J. The amount of energy transformed is also 1000J. Delta E denotes energy change. The force of gravity does 1000J of work to convert 1000J of potential energy into 1000J of kinetic energy. The amount of work done is equal to the amount of energy changed.

When the object strikes the ground, the kinetic energy is converted into sound and heat energy. What is the sum of heat and sound energy produced in this incident? Yes, it’s 1000J. A sum of 1000J of sound and heat energy is produced. When the object stopped, all the 1000J of kinetic energy is converted into heat and sound energy.

How much work is done to stop the object? Yes, it’s 1000J, too. This is because the amount of work done is equal to the amount of energy changed. Since the energy changed is 1000J, hence the work done must also be 1000J.

This is a very useful work-energy relationship. Amount of work done is equal to the amount of energy changed. We can see that we don’t need to use the equation W = F x s to find the work done. We just need to know how much energy is transformed, and then we will straight away know how much work is done.

Let’s see this example. An object of mass 2kg is moving at an initial velocity of 2 m/s. A force F exerts on the object, causing it accelerates from 2 m/s to 5 m/s. Find the work done by the force.

As usual, let’s list down all the information that we have. The mass, m = 2kg. The initial velocity, u = 2 m/s. The final velocity, v = 5 m/s. We are asked to find the work done, W.

Well, work done, W = F x s. But we don’t know the F and s. Can we find F by using the equation F =ma? We have m, but we don’t have a. Can we find a by using the equation v = u + at? We have v, we have u, but we don’t have t. So, obviously, we can’t find the work done by using the equation W = F x s.

So, what other equation we can use? Well, we can find the work done by using the work-energy relationship. In previous slide, we have learned that amount of work done is equal to the amount of energy changed. In this case, we can find the kinetic energy change. The kinetic energy change is equal to the final kinetic energy – the initial kinetic energy, which is equal to ½ mv^{2} – ½ mu^{2}. Plug in the value of m, v and u. The energy change is equal to 21J. Remember, amount of work done equal to amount of energy changed. To make the kinetic energy increases by 21J, we need to do 21J of work. Therefore, the work done is 21J.

This is a typical example of using the work-energy relationship to find the amount of work done.