Work, power and efficiency - CCEA

• Work is said to be done when energy changes from one form to another.

• The amount of work can be calculated by the equation work = force × distance, work is measured in joules (J), the force in newtons (N) and the distance in metres (m).

• Power is the amount of energy transferred in one second or the amount of work done in one second.

• Power is measured in watts (W) so 1 W = 1 joule per second (1 W = 1 J/s).

• power = \(\frac{\text{energy transferred}}{\text{time taken}}\); power = \(\frac{\text{work done}}{\text{time taken}}\)

• Efficiency is a measure of how much of the input energy to a process or device appears as useful output energy.

• Efficiency = \(\frac{\text{useful output energy}}{\text{total input energy}}\), efficiency is given as a decimal or a percentage but has no units.

Here is the equation that relates to work done, force applied, and distance moved in the direction of the force.

\(\text{Work done} = {\text{Force}}\times{\text{distance}}\)

\(\text{W} = {\text{F}}\times{\text{d}}\)

Where

W is work done measured in joules, J

F is force measured in newtons, N

d is distance measured in metres, m

In the example above, 10 N is applied to move the box 2 m.

Work done W = Fd

F = 10 N

d = 2 m

Work done = 10 × 2 = 20 J

The work done pushing the box 2 m is 20 J.

Equations for calculating work

Energy transferred = work done, and so we can also say:

Power is the amount of work done in one second.

Power is measured in watts (W).

How to calculate power

Power can be calculated using either of these equations:

power = \(\frac{\text{work done}}{\text{time taken}}\)

work = power x time taken

time taken = \(\frac{\text{work done}}{\text{power}}\)

Apparatus

12 V electric motor, power pack, connecting leads, string, known weight, stop clock, felt tip pen, two cardboard pointers.

Method

The power pack is connected to the 12 V motor using the connecting leads.

The 12 V motor is attached to the known weight by a length of string wound round the spool.

Record the weight in newtons in a table.

Two pointers A and B are attached to the bench close to the string.

Measure the distance between the two markers with a metre rule and recorded in metres.

A suitable distance is between 1 m and 2 m.

A felt tip pen is used to mark a clearly visible dot on the string.

When the motor is switched on the weight is lifted at a constant speed.

Start the stop clock when the dot passes pointer A.

Stop the stop clock again when the dot passes pointer B.

Record the time in seconds in the table.

Repeat this process twice more and calculate the average time taken.

This helps to reduce error due to reaction time when starting and stopping the stop clock.

The work done in lifting the weight is calculated using:

work done W = Fd

The power of the motor is then calculated using:

power = \(\frac{\text{work done}}{\text{time taken}}\)

The process can be repeated for a range of increasing weights.

To ensure that this is a fair test the height the weight is raised (i.e. the distance between the pointers) and the voltage are kept constant.

Results

1. Measure the persons mass in kg using bathroom scales.

2. Convert it to weight using W = mg. This equals the upward force that will move up the stairs.

3. Measure the height of 1 step using the 50 cm rule. Record in m in a suitable table. Repeat for two more steps and calculate the average height in m.

4. Count the number of steps and record.

5. Calculate the vertical height = number of steps x average height of 1 step.

6. Calculate the work done = force x vertical height.

7. Time the person running the stairs. Record in the table. After a rest repeat the experiment and calculate average time.

8. Calculate the persons average power using: power = \(\frac{work~done}{time~taken}\).

Key fact

• Efficiency = \(\frac{\text{useful output energy}}{\text{total input energy}}\)

Efficiency = \(\frac{\text{useful output energy}}{\text{total input energy}}\)

Useful output energy = 10 J

Total input energy = 100 J

efficiency = \(\frac{\text{10}}{\text{100}}\)

efficiency = 0.1

This means that 0.1 (or 10 %) of the electrical energy supplied is transferred as light energy.

The light bulb is not very efficient since most of the energy supplied is not transferred usefully.

Most of the energy will have been dissipated as heat energy.

This is because light bulbs become very hot when they are switched on.

Another way of thinking about this is that for every £100 spent on lighting, £10 is spent on the light and £90 is wasted heating the surroundings.

The efficiency of a device can never be greater than 1 otherwise energy would be created, and the Principle of Conservation of Energy violated.

Occasionally power is shown in W instead of energy in J. The equation for efficiency is similar – just substitute power for energy:

efficiency = \(\frac {useful~output~power}{total~input~power}\)

Efficiency = \(\frac{\text{useful output energy}}{\text{total input energy}}\)

Useful output energy = 75 J

Total input energy = 100 J

efficiency = \(\frac{\text{75}}{\text{100}}\)

efficiency = 0.75

This means that 0.75 (or 75 per cent) of the electrical energy supplied is transferred as light energy.

For every £100 spent on lighting, £75 is spent on the light and only £25 is wasted heating the surroundings.

The LED lamp is much more efficient than the filament lamp and can help save money.