# Interactive textbook: Work energy and Power

### So far so good Physics is all about understanding the way the matter in the universe behaves. We do this by constructing models that can be used to make predictions. Before constructing models we defined the quantities needed to describe the events we observe. We started with length and time to put numbers to events we saw taking place at different positions. The observation of motion led to the need for displacement and velocity and the interaction between two particles led to acceleration, force, mass, momentum and impulse. We also observed that these quantities were related in ways that we expressed mathematically in the equations for constant acceleration, Newton's laws of motion and the principle of conservation of momentum.

Can we now predict the outcome of any interaction between two particles? Let's try.

Consider two balls travelling towards each other as shown. We know that momentum will be conserved, the momentum before the collision is zero so the momentum after will be zero but this does not have a unique solution. In fact there are an infinite number of solutions, such as:

All of these examples obey the principle of conservation of momentum but we are missing something. The impulse of each ball in the first example is twice as big as the last so the force on each ball is greater. What enables the balls to push harder? The answer is energy... but first let's look at the process of pushing.

### Work and energy When we observe collisions between two bodies it looks like the collision takes no time at all. This can't be true as it would mean infinite acceleration and therefore infinite force. When body A pushes body B there is movement in the direction of the force.

We'll come back to the example above later. For now, we will look at a different but familiar example of a box on a frictionless surface pulled by a ball hanging on a rope.

The box is being accelerated by an unbalanced force, the tension. The change in momentum of the box depends on two things: the size of the force and the distance moved. Let's define a new quantity that combines these factors, work:

work done = force x distance moved in the direction of the force

Once a box is moving it has the ability to do work on something else.

We call the ability to do work energy:

energy = the ability to do work

When body A does work on body B, energy is transferred from A to B.

There are two reasons why one body is able to do work on another:

1. Because it is moving
2. Because of its position

The hanging ball was able to do work on the block because of its position above the Earth; it had potential energy. It gained kinetic energy when it moved faster.

If we look more closely at the colliding boxes we can deduce - because the forces on the two boxes are equal and opposite and the distance moved by each is the same - that the work done by the brown box is equal to the work done on the yellow one. The energy lost by the brown is equal to the energy gained by the yellow. Energy is conserved. This gives us the law of conservation of energy:

Energy can neither be created or destroyed. It can only be converted from one form to another.

This is probably the most important law in physics and enables us to solve the two ball problem.

Before going into work and energy in more detail let's complete our analysis of the hanging ball example. The work done by gravity on the ball is much larger than the work done on the box. Where does this extra energy go?! No need to worry. The ball is also accelerating and gains kinetic energy.

A ball is dropped from a high building and falls to the ground.

Where does it has most energy?

Energy is conserved

#### Calculating work

Let's consider the simplest case of a constant force acting in the direction of motion.

$$W=Fd$$

The unit of both work and energy is $$\text{N m}$$. This is the same as a Joule, $$\text{J}$$.

• $$W$$ is work done in $$\text{J}$$
• $$F$$ is force in $$\text{N}$$
• $$d$$ is the distance moved in the direction of motion in $$\text{m}$$ (sometimes we instead use $$s$$ to represent displacement)

Both force and displacement are vectors, so is work a vector or scalar? There are two ways of multiplying vectors $$\mathbf{A}$$ and $$\mathbf{B}$$:

1. $$\mathbf{AB}\cosθ$$ (dot product) is a scalar
2. $$\mathbf{AB}\sinθ$$ (cross product) is a vector

For work we use the dot product, so work is therefore a scalar (as is energy).

If the force acts at an angle to the direction of motion then we use the component of force in the direction of motion:

$$W = F\cosθ \times d$$

In this case the work done by the force is zero (because $$\cos90=0$$):

And in this case work is negative:

Since work is a scalar the sign has nothing to do with direction. If the work done on the body is positive the body gains energy. If it is negative it loses energy. If the body does positive work, it transfers energy to the body it is doing work on. If the work done by the body is negative, it gains energy from the body.

In this example F = 10 N and θ = 60°

How much work is done per meter?

Work in 1 m = 10 x cos60 x 1

#### Graphical treatment

Work done can be found from the area under a force vs distance graph. In this example the work done is $$8\text{ J}$$.

This is a force - distance graph for a non uniform force. How much work is done in 2m?

Area = 1/2 x 2 x 2 = 2 J

#### Changing direction

All curves (when zoomed in enough) can be approximated as a lot of short straight sections. To calculate the work done along a curved path we can find the work done for each short section then add them all up to find the total. Here a constant force pushes the box around a curved path. Note that there are other hidden forces that cause the box to change direction but these forces are perpendicular to the motion so don't do work.

This motion can be split into many short approximately straight sections. In this section, $$\Delta W=F\Delta d$$

The work done for the whole motion is the sum of all the small steps: $$W=F\times\text{path length}$$

When a body travels in a circular path at constant speed there is a force acting towards the centre of the circle. Is work done by this force?

The force is perpendicular to the direction of motion

### Kinetic energy Kinetic energy is the energy a body has due to its motion. To have gained kinetic energy, work must have been done on the body. Because energy is conserved the work done in accelerating the body to a given velocity will equal its kinetic energy.

From the definition of work: $$W=Fs$$

From Newton's 2nd law: $$F=ma$$

Combining these equations: $$W=mas$$

The acceleration is constant so applying the 'suvat' equation: $$v^2=u^2+2as\Rightarrow as={1\over2}v^2$$

And substituting this into the equation for work gives:

$$W={1\over 2}mv^2$$

$$E_\text{k}={1\over2}mv^2$$ ### Potential energy Potential energy is the energy a body has due to its position. The name is a bit misleading as it means the 'potential to do work' but a body with kinetic energy has the potential to do work too. It is better to think of the 'p' as standing for position. There are many reasons why the position of a body might cause it to have potential energy. Here we will consider the position above the Earth, with the energy type known as gravitational potential energy.

#### Gravitational potential energy

The potential of a body in a given position is equal to the work done in putting it there from a position of zero potential energy. For a body to really have no potential energy, it would have to be an infinite distance from the Earth. This makes things complicated (!) so we will take the surface of the Earth (the ground) to be zero. The potential energy of a body above the ground is then the work done in lifting it from the ground to that position.

Note that the forces on the body are balanced so it is travelling with constant velocity. If they weren't balanced it would also be accelerating so would gain kinetic energy.

Applying the definition of work: $$W=Fh$$

Since the forces on the ball are balanced: $$F=mg$$

$$E_\text{p}=mgh$$

#### Elastic potential energy

Elastic potential energy is related to springs and elastic strings. The force required to stretch a spring gets bigger the longer it gets.

Hooke's law states that the force required to extend a spring is directly proportional to the extension of the spring. A graph of force vs extension will therefore be a straight line. The gradient of this line is the spring constant, the constant of proportionality between force and extension:

$$F=kx$$

For this example, $$k={F\over x}=0.8\text{ N cm}^{-1}=0.08 \text{ N m}^{-1}$$

Stretching the spring requires work to be done. This increases the elastic potential energy of the spring. The work done can be found from the area under the graph:

$$W=\text{area under graph}={1\over 2}Fx$$

Given $$F=kx$$:

$$W={1\over 2}kx^2$$

$$E_\text{p}={1\over 2}kx^2$$

A 2 kg body is held at a height of 5 m above the ground and released.

What is the PE of the body?

PE = mgh = 2 x 10 x 5

A 2 kg body is held at a height of 5 m above the ground and released.

What is its KE half way down?

Loss of PE = gain in KE = 2 x 10 x 2.5

A 2 kg body is held at a height of 5 m above the ground and released.

What is its speed just before it hits the ground?

Final KE = initial PE = 1/2mv2 =100

v = √(2 x 100/2)

### Energy changes

The total energy of an isolated system is constant. We can use this fact to model the motion of bodies. This is often a simpler way of solving problems than using the equations of motion.

#### Ball thrown upwards

When a ball is thrown upwards it starts with kinetic energy. This is converted to potential energy as the ball rises as work is being done against the force of gravity. At the top the ball has no kinetic energy but has potential energy.

loss of $$E_\text{k}=$$ gain in $$E_\text{p}$$

$${1\over 2}mv^2=mgh$$

$$\Rightarrow v=\sqrt{2gh}$$

This can also be found using the 'suvat' equations (provided that acceleration is constant):

$$v^2=u^2+2as$$

$$v=\sqrt{2gh}$$

#### Mass fired from a spring

A ball is fired horizontally from a toy gun.

Elastic potential in the spring is converted to the kinetic energy of the ball.

During the projectile motion the horizontal velocity stays the same but the vertical component of velocity increases. This results in an increase in speed. Gravitational potential energy is being converted to kinetic energy.

#### Friction

When a box slides along a surface the force of friction acts in the opposite direction to the motion. The block does work against friction so energy is lost. Where does this energy go? To explain this we need too consider the atoms that make up the box.

The energy goes to increase the kinetic and potential energy of the atoms. This combination of energies is called internal or thermal energy.

It might not be obvious why the potential energy increases. The atoms are held together by an interatomic force so work has to be done to move them apart. You will learn all about this in the section on Thermal Physics.

#### Box sliding down a slope

When a box slides down a slope, potential energy is converted to kinetic energy. If the surface is frictionless the original potential energy is equal to the final kinetic energy.

$${1\over2}mv^2=mgh$$

$$v=\sqrt{2gh}$$

If there is friction on the ramp then some work is done against this. Not all the potential energy is converted to kinetic energy; some becomes internal energy:

$$mgh={1\over 2}mv^2+W_\text{friction}$$

$$mgh={1\over 2}mv^2+\mu_\text{d}N\times \text{distance down slope}$$

A 2 kg body slides down a slope. The height of the slope is 4 m and the velocity of the body at the bottom is 5 ms-1.

How much work is done against friction?

Initial PE = 80 J

Final KE = 25 J

change in PE = gain in KE + work done against friction

80 = 25 + W

### Power

Power is the rate at which work is done. It is a scalar quantity and has units $$\text{J s}^{-1}$$ which is the same as the Watt ($$\text{W}$$).

High power means that a lot of work is done or energy is transferred in a short time. Anyone can do a lot of work but not everyone can do it in a short time. Lifting a hundred bags of sugar from the ground into a trolley is something that all of us could do... but we couldn't all do it in 10 seconds. We have enough energy but not enough power. To do this job in 10 seconds would require us to either lift each kilogram bag of sugar very quickly or lift them all at once.

#### Power at constant velocity

A powerful car can accelerate from rest to $$100\text{ km h}^{-1}$$ in a very short time and has a high top speed. While accelerating, the kinetic energy of the car is increasing. The rate at which kinetic energy increases depends on the power of the engine. At top speed the forces on the car are balanced; the forward force is equal to drag. To move quickly the rate of doing work against drag must be high.

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

$$P={Fd\over t}=F{d\over t}=Fv$$

This equation can only be used if the velocity is constant.

A truck is powerful but can't achieve high accelerations, nor does it have a high top speed. Instead, trucks move heavy loads at a reasonable speed.

A car travelling at constant velocity of 20 m s-1 experiences a drag force of 1000 N.

What is the power of the car

Engine force = drag since velocity is constant

#### Energy flow

These examples are quite different to all of the examples involving balls and boxes used in this unit so far. We can't really explain what is happening using simple physics. Work is done when body A pushes body B - but where is the body that is pushing the car? To answer this question we have to look at the situation on an atomic level.

A vehicle uses petrol to produce movement. Petrol is made up of atoms arranged in groups called molecules. To arrange the atoms in this way required work to be done. This involves chemistry, but all we need to know is that if work was done on the atoms to put them in the molecule then they have potential energy. When petrol is burnt the atoms are rearranged into other molecules that have less potential energy.

Since energy is conserved, the molecules gain kinetic energy (move faster). The fast-moving molecules hit the pistons in the engine. These collisions transfer energy to the piston, causing it to move. They also cause the atoms in the metal of the piston to move faster, which increases the internal energy of the piston. The moving pistons turn the wheels of the car, increasing the kinetic energy of the car.

To model this on a particle level is rather complicated but we can explain where the energy comes from using a simple flow diagram. This is a "systems" approach often used when we need an overview without going into the details of what is happening to each individual particle. ### Efficiency

In the example of a car engine we can see that not all of the energy from the fuel was converted to the kinetic energy energy of the car. The process was not 100% efficient. It's time to define efficiency:

$$\text{efficiency}={\text{useful work output}\over\text{total energy input}}={\text{useful power output}\over\text{total power input}}$$

For the car, $$\text{efficiency}={E_\text{k, car}\over E_\text{fuel}}$$

#### Alternative energy units Newspaper articles about power production are not always written by journalists with a complete understanding of the difference between energy and power or their units. When dealing with the generation of electrical energy the unit kilowatt-hour is often used. This is the amount of energy produced by a $$1\text{ kW}$$ generator in $$1\text{ hour}$$:

$$1\text{ kW h}=1000\text{ W}\times3600\text{ s}=3.6\times10^6\text{ J}$$

Note that a common mistake is to write $$\text{kW}/\text{h}$$, which is the rate of change of power. It probably comes from confusion with $$\text{km h}^{-1}$$. To a non-scientist "kilowatt per hour" might sound better than "kilowatt-hour".

Another mistake is to make statements like "The power station will produce $$12\text{ GWh}$$" (where $$1\text{GW}=10^9\text{ W}$$). This tells us nothing about the power station as any power station can produce $$12\text{ GWh}$$ if given enough time. The statement should be "The power station will produce $$12\text{ GW}$$ of power" or "The power station will produce $$12\text{ GWh}$$ of energy every hour". A car travelling at constant velocity of 20 m s-1 experiences a drag force of 1000 N. The car uses 1 ml of fuel per second.

If the energy density of the fuel is 10 MJ l-1 what is the efficiency of the car?

Energy used per second = 0.001 x 10000000 = 100000 = 100 KJ

Power in = 100 kW

Efficiency = Power out/power in x 100%  = 20/100 x 100% = 20%

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