Forms of energy, conservation and work done

​​In a nutshell

Energy comes in many different forms and can be transferred from one to another. Work done by a force is the energy it transfers and this can be calculated both parallel, perpendicular and at an angle to the plane. The principle of the conservation of energy prevents energy from being created or destroyed.

Equations

Variable definitions

Note: A joule is equal to one Newton metre. $$1\,J$$ is the work done when a force of $$1\,N$$ moves its point of application $$1\,m$$ in the direction of the force.

Forms of energy

Energy is the capacity for doing work and is found in many different forms. Potential energy refers to any type of energy that can be stored such as gravitational or elastic energy. Energy of one form can be transferred into another in an energy transfer process.

Some different forms of energy are found in the table below:

The energy transfer from a source is equal to the work done. This is not necessarily the total energy.

Lifting a bag from a chair to a table higher than it will involve an energy transfer. The gravitational potential of the bag will increase as it is lifted but since it possessed some gravitational potential energy at the start, the work done is just the increase in the gravitational potential energy.

There is an energy transfer taking place when you eat food. Chemical energy in the food can be transferred to thermal energy as your body heats up when digesting food, as well as into mechanical energy through muscle movements.

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Conservation of energy

The principle of the conservation of energy states that the total energy of a closed system remains constant. Energy cannot be created or destroyed, only transferred from one form to another.

When energy transfers take place some energy may be transferred into non-useful forms i.e. some energy is 'lost', but the total final energy is always equal to the total initial energy.

Example

A piece of coal contains has $$100\%$$​ of its initial energy in the form of chemical energy. This coal is burned on a BBQ. State what energy transfers occur as the coal is burned.

As the coal is burned, the stored chemical energy is transferred into other forms such as thermal energy. There will also be some energy transferred into light and sound energy.

Note: The thermal energy released by coal is used in power stations to produce electricity by driving a turbine that spins an electrical generator.

Example

An electrical lamp is plugged into the mains. When switched on, electrical energy supplied to the lamp undergoes energy transfers into light, heat and sound energy. If $$15\%$$ of the initial energy is converted into light energy and $$1\%$$ into sound, what percentage of the initial energy is converted into heat energy?

Tip: Assume that all the electrical energy is transferred into light, heat or sound energy forms only.

Firstly, write down the key information from the question:

Initially, $$100\%$$ of the energy is electrical energy.​

Next, write down the known energy transfers given in the question:

​$$15\%$$ of this energy is converted to light energy and $$1\%$$ into sound energy.

Now calculate how much energy is not converted into light and sound energy:

$$100\% - 15\% -1\% = 84\%$$​​

The percentage of energy that must be converted into heat energy is $$\underline{84\%}$$.

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Calculating work done

When the applied force and the displacement act in the same direction, the work done by the force can be calculated by:

​$$work \ done = force \ \times \ distance \ (moved \ in \ the \ direction \ of \ the \ force) \\ W = Fx$$​​

Note: This equation only applies when the direction of the motion are parallel

Example

A ball weighing $$150\,g$$ falls $$2.5\,m$$ to the ground. Calculate how much work is done as this occurs to 2 s.f.

Firstly, write down the given values and convert units where needed:

$$m = 0.150\,kg \\ x = 2.5\,m$$​​

Secondly, write down the equation needed and calculate the force:

$$W = Fx$$​​

Tip: The force on the ball is its weight, $$mg$$, where $$g = 9.81 \,ms^{-1}$$​.

$$F = mg = 0.150\times 9.81 = 1.47\,N$$​​

Next, substitute values into the equation:

$$W = 1.47 \times 2.5 = 3.675\,J$$​​

Calculate the final answer and include units:

$$W = 3.7\,J$$​​

The work done by the force as the ball falls in $$\underline{3.7\,J}$$.

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Tip: Before starting a question, consider the direction the force is applied in and the direction of the displacement!

Forces at an angle

When the direction of movement differs from the direction the force is applied in, you must calculate the component of the force that is acting in the direction of the movement. Only this component of the force is causing the displacement, and therefore, is doing work.

The component of the force in the direction of motion is found by:

​$$F\cos{\theta}$$​​

Therefore, the work done is:

$$W = Fx\cos{\theta}$$​​

To use this equation, you need to know the angle between the force and the direction of motion, $$\theta$$.

Example

A boy pulls a sledge along the ground. He pulls with a force of $$72\,N$$ along a $$23\,m$$ path. He holds the string attached to the sledge at an angle of $$77\degree$$ to the horizontal ground. How much work is done by the boy as he pulls the sledge to 2 s.f. 

Firstly, write out the figures given in the question:

$$F = 72\,N \\ x = 23\,m \\ \theta = 77\degree$$​​

Then, write the equation you need to use:

$$W = Fx\cos{\theta}$$​​

Note: In this question, since the direction of movement is horizontal along the ground, only the component of the force that is in this direction needs to be considered. Therefore, the above equation is used.

Substitute the values into the equation:

$$W = 72 \times 23 \times \cos{\theta} = 372.5...\,J$$​​

Calculate the final answer with appropriate units:

$$W = 370\,J$$​​

The total work done by the boy in horizontal direction is $$\underline{370\,J}$$​.

If the angle of the string was $$0\degree$$ with respect to the horizontal case, then the equation would simplify to the previous case ($$W = Fx$$) as $$\cos{\theta} = \cos{0} = 1$$.

Perpendicular forces

Sometimes, the force may be applied to an object at $$90\degree$$ to the direction of motion. To calculate the work done you must only consider the component of the force that is in the same direction as the movement of the object. In this case, the applied force does not cause any movement in the direction of motion and therefore does not do any work on the object as $$cos(90) = 0$$​.