Work, Energy and Power

A LevelAS LevelAQA

Work, Energy and Power

One way in which energy can be transferred is when work is done. Work can be done when pushing an object, for example. Power is the rate at which this energy is transferred when work is done.

Work Done

Work done is defined as “the amount of energy transferred when a force causes an object to move”. The equation for work done is:

W=F×sW = F times s

  • WW is the work done in joules (J)left(text{J}right).
  • FF is the force in Newtons (N)left(text{N}right).
  • ss is the distance in metres (m)left(text{m}right).

We can see from this equation that an alternative unit for energy is Nmtext{Nm}.

When an object is moved over a certain distance, work is done to overcome the forces of friction, air resistance and any other drag forces. For this calculation to be applied, the measurement of distance must be parallel to the direction the force is applied.

Example: If an object is moved 20m20 : text{m} with a constant force of 1500N1500 : text{N}, how much work is done?

W=F×sW=1500×20=30000Jbegin{aligned} W &= F times s W &= textcolor{bd0000}{1500} times textcolor{7cb447}{20} = boldsymbol{30000} : textbf{J}end{aligned}

 

However, often the force being applied is not parallel to the direction of motion. To calculate the work done, we need to resolve the force into its components to find the component in the direction of motion. This gives us 2 new equations:

W=Fcosθ×sW = F cos theta times s

W=Fsinθ×sW = F sin theta times s

 

Example: Calculate the work done moving the object below. 

[2 marks]

W=Fcosθ×sW = F cos theta times s

W=1500×cos25×20boldsymbol{W = 1500 times cos 25 times 20 }

W=27190Jboldsymbol{W = 27190} : textbf{J}

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Power

Power and work done are related.

Power is the rate of energy transfer from one form of energy to another or from one object to another. It is also the rate of work done. Therefore it can defined using the equation:

P=ΔWΔtP = dfrac{Delta W}{Delta t}

  • PP is the power in watts (W)left(text{W}right).
  • ΔWDelta W is the work done in joules (J)left(text{J}right).
  • ΔtDelta t is the time taken for the work to be done, in seconds (s)left(text{s}right).

Example: A cyclist does 12000J12000 : text{J} of work on a bicycle to move for 3030 seconds. What is the cyclist’s power?

P=ΔWΔtP=1200030=400Wbegin{aligned} P &= dfrac{Delta W}{Delta t} P &= dfrac{12000}{30} = boldsymbol{400} : textbf{W} end{aligned}

 

Alternatively, we can use another equation to calculate power if we know the force and velocity of a moving object:

P=F×vP = F times v

  • vv is the velocity of the object in metres per second (ms1)left(ms^{-1}right).

This is because:

P=ΔWΔt=FsΔt=FvP = dfrac{Delta W}{Delta t} = dfrac{Fs}{Delta t} = Fv

 

Example: A man is moving house and carries a box, using a force of 200N.200 : text{N}. He moves the box 2020 metres in 1010 seconds. Calculate the power.

[2 marks]

V=st=2010=2ms1V = dfrac{s}{t} = dfrac{20}{10} = boldsymbol{2} : textbf{ms}^{-1}

P=F×v=200×2=400WP = F times v = 200 times 2 = boldsymbol{400} : textbf{W}

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Force-Displacement Graphs

A force-displacement graph can be used to show how the force applied to an object changes over the displacement it has been moved. This is useful as forces are not always constant.

For example, a person pushing an object along the floor will need to vary their force to overcome changes in friction with the floor surface.

Work done can be calculated using a force-displacement graph by calculating the area under the graph. This is the equivalent of Work=Force×Displacementtext{Work} = text{Force} times text{Displacement}.

Example: The graph below shows how the force of carrying an object changed with displacement. Calculate the work done carrying the object 33 metres.

[3 marks]

Firstly, we need to split the graph into sections so that we can calculate the area of each section. This is because work done is equal to the area under a force-displacement graph.

Area of rectangle=16×0.5=8Jtext{Area of rectangle} = 16 times 0.5 = boldsymbol{8} : textbf{J}

Area of triangle=12×16×2.5=20Jtext{Area of triangle} = dfrac{1}{2} times 16 times 2.5 = boldsymbol{20} textbf{J}

Total work done=8+20=28Jtext{Total work done} = 8 + 20 = boldsymbol{28} : textbf{J}

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Efficiency

The efficiency of a system or a device is a measure of its ability to transfer the input energy into useful output energy. This can be expressed as a decimal between 0 and 1 or as a percentage from 0 to 100%. The equation for calculating efficiency as a percentage is:

Efficiency=Useful energy outputTotal energy input×100text{Efficiency} = dfrac{text{Useful energy output}}{text{Total energy input}} times 100

Or we can also use:

Efficiency=Useful power outputTotal power input×100text{Efficiency} = dfrac{text{Useful power output}}{text{Total power input}} times 100

No device or system can be 100%100 % efficient but the more efficient a device, the less energy is wasted. If you wanted to calculate efficiency as a decimal instead of a percentage, simply don’t multiply by 100100.

Example: A television requires a power input of 200W200 : text{W}. It transfers this power into 75W75 : text{W} of sound, 75W75 : text{W} of light and 50W50 : text{W} of heat. Calculate the efficiency of the TV.

[2 marks]

Total input energy=200wtext{Total input energy} = 200 : text{w}

Total useful power output=75+75=150Wtext{Total useful power output} = 75 + 75 = 150 : text{W}

Efficiency=Useful power outputTotal power input×100text{Efficiency} = dfrac{text{Useful power output}}{text{Total power input}} times 100

Efficiency=150200×100=75%text{Efficiency} = boldsymbol{dfrac{150}{200} times 100 = 75 % }

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Work, Energy and Power Example Questions

Question 1: A wind turbine does 3.1×107J3.1 times 10^7 : text{J} of work converting kinetic energy into electrical energy in one day. What is the power output of the wind turbine?

[2 marks]

A LevelAS Level AQA

Firstly, convert 11 day to seconds:

1×24×60×60=86400s1 times 24 times 60 times 60 = boldsymbol{86400} : textbf{s}

P=ΔWΔtP=3.1×107J86400=360Wbegin{aligned} P &= dfrac{Delta W}{Delta t} P &= dfrac{3.1 times 10^7 : text{J}}{86400} = boldsymbol{360} : textbf{W} end{aligned}

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Question 2: A winch pulls an object with a force of 2700N2700 : text{N} at 60°60 degree to the horizontal. It moves 150m150 : text{m} horizontally. Calculate the work done.

[3 marks]

A LevelAS Level AQA

A diagram can be drawn to represent the information in the question:

 

W=Fcosθ×sW=2700cos60×150W=202500Jor0.2MWbegin{aligned} W &= F cos theta times s W &= 2700 cos 60 times 150 W &= boldsymbol{202500} : textbf{J} : text{or} : boldsymbol{0.2} : textbf{MW} end{aligned}

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Question 3: A bicycle motor produces 5kW5 : text{kW} of power. The motor is 85%85 % efficient at converting its power into kinetic energy. What is the maximum useful power output of the bicycle?

[3 marks]

A LevelAS Level AQA

Efficiency=Useful power outputTotal power input×100Useful Power output=Efficiency100×Total power outputUseful Power output=85100×5000Useful Power output=4250Wbegin{aligned} textbf{Efficiency} &boldsymbol{= dfrac{textbf{Useful power output}}{textbf{Total power input}} times 100} text{Useful Power output} &= dfrac{text{Efficiency}}{100} times text{Total power output} text{Useful Power output} &= boldsymbol{dfrac{85 }{100} times 5000} text{Useful Power output} &= boldsymbol{4250} : textbf{W} end{aligned}

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Specification Points Covered

AQA A Level:

  • 3.4.1.7 Work, Energy and Power