JR/2008 Work & Energy If an object moves due to the action of an applied force the force is said to have done WORK on the object. Work is the product of.

JR/2008

Work & Energy

If an object moves due to the action of an applied force the force is said to have done WORK on the object.

Work is the product of force applied and the distance moved in the direction of the force.

WORK DONE = FORCE x DISTANCE

The unit of work can be found from

Force = Newtons

Distance = metres

Work is therefore measured in Newtons x metres or Newton . metres

However:- Work is usually defined in units of Joules (J)

JR/2008

Work & Energy

The Joule is defined as:-

The amount of work done when a force of 1 Newton acts for a distance of 1 metre.

1 Joule = 1 Newton metre

1 J = 1 N.mIf a graph is plotted of Force ( vertical y axis) against :

distance ( horizontal x axis) the force/distance graph is given.

This graph is known as a WORK DIAGRAM

The area beneath the graph equals the work done

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Work & Energy

The work diagram

a. Constant force

b. Changing force

Force

F

Work done = area of graph

WD = F.s (Nm or J)

Work done = area of graph

WD = ½ F.s (Nm or J)

Distance (s)

Force

F

Distance (s)

JR/2008

Work & Energy

Ex1 A spring is stretched 25mm by a force of 50N. Draw a diagram of the work done over the next 50mm of stretch and from this find the work done in that 50mm and the average force applied

Solution:-

Spring stiffness = 50N/25mm

= 2N/mm = 2kN/m

The next 50mm needs 50x 2N/mm

= 100 N

Work done = area of graph

WD = ½ (50 + 150) x 50 x10-3

= 5 Nm or 5 J

Average force = (50 + 150)/2

= 100N

Force

N

Distance (mm)7525

50

150

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Work & Energy

Examples:

1. Calculate the work done in lifting a mass of 0.5 tonne through a vertical distance of 10m

Soln.

Work = force x distance

In this case the work is done against gravity – thus force = weight (mg).

When:- mass = 0.5 tonne = 500 kg, g = 9.81, distance = height 10m

Work = mass x g x height = 500 kg x 9.81 x 10 = 49050 J = 49.05 kJ

JR/2008

Work & Energy

Examples:

2. A sledge is pulled along horizontal ground at a constant speed by a rope which makes an angle of 25° with the horizontal. If the tension in the rope remains constant at 100N calculate the energy supplied in moving the sledge 1 km.

Soln.

Work = force x distance

In this case the work is done in the horizontal direction. To find the force in this direction we must resolve the tension (force) in the pulling rope.

Horizontal pull = T cos 25° = 100 cos25°

= 90.63 N

Work = force x distance = 90.63 x 1000m

Work = 90.63 kJ (in the horizontal direction)

25°

Tension T

Horizontal pull = T cos 25 °

JR/2008

Examples:

3. A force acts on a body of mass 15kg, which is initially at rest, and imparts a constant acceleration of 2 ms-2 for 10 seconds. Calculate the energy transfer (work done ) to the body.

Soln.

Work done = Energy used = force x distance

In this case the equations of motion must be used to find distance and force.u = 0 ms-1

v = ?

a = 2 ms-2

s = ?

t = 10 seconds

Work & Energy

From s = ut + ½ at2 distance s = ½ x 2 x 102 = 100m

From N2 F = ma, Force = 15 kg x 2 ms-2 = 30N

Work done = Energy transferred = 30N x 100m = 3000 J

Energy transfer = 3 kJ

JR/2008

Work & Energy

Energy is defined as the ability to do work

WORK DONE = ENERGY USED

The unit of ENERGY is as work:-

Newtons x metres or Newton . Metres or Joules (J)

Energy is given up when work is done

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Work & Energy

Typical forms of energy are:

Mechanical

Electrical

Heat

Chemical

Nuclear

Light

Sound

Energy may be converted from one form into another

In doing so its value remains the same

JR/2008

Work & Energy

Conservation of Energy:- the principle states that:-

ENERGY CANNOT BE CREATED OR DESTROYED

Typical conversions include:-

Mechanical into electrical by a generator

Electrical into mechanical by a motor

Heat into mechanical by a steam turbine

Chemical to heat by combustion

Nuclear to electrical via a steam turbine & generator

Sound to electrical by a microphone

Many other conversions are regularly seen, try to discover an area where energy is simply ‘lost’ or created from nothing.

JR/2008

Work & Energy

The conversion of energy is never simple – and the conversion is never 100 % efficient, losses will always occur.

A loss of Energy does not mean a reduction in the overall quantity but a conversion into a form which is not required.

A simple example is the petrol engined motor car.

The fuel (chemical energy) is intended to provide motion.

In fact only 25% of the energy within the fuel provides motion, the remaining 75% is converted to heat, vibration, sound etc.

The car is said to be 25% efficient.

Great efforts are made in engineering to try and improve the efficiency of energy conversion.

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Work & Energy

The conversion process can be simplified with a block diagram:-

Input Energy = Output Energy + ‘Lost’ Energy

EFFICIENCY = OUTPUT x100%

INPUT

LOSSES

PROCESS OUTPUTINPUT

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Work & Energy

Using the equation:- EFFICIENCY = OUTPUT x100%

INPUT

It can be seen that output will always be < input thus efficiency will always be < 100%

The symbol used for efficiency is η (eta) and is given as a percentage

Ex1. An electric motor takes 5000 J from the mains supply. If the output shaft of the motor provides 3600J find the efficiency of the motor.

Soln. Efficiency = Output/Input x 100%

η = (3600 J ÷ 5000 J) x 100% = 72%

the 28% lost energy is probably converted to heat

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Work & Energy

Ex2. A petrol engine has an efficiency of 32% and delivers 12.5 kJ/second at the crankshaft. What is the energy input to the engine.

Soln.

Efficiency = output/input

32/100 = 12.5 kJ/s ÷ Input

Input = 12.5 ÷ 0.32 kJ/s = 39.0625 kJ/s

Engine

η = 32%Input 12.5 kJ/s

JR/2008

Work & Energy

Ex3. A motor has an efficiency of 75% and drives a gearbox with an efficiency of 87%. If the output from the gearbox must be 20.76 kJ/s

calculate the motor input in kJ/s

Soln.

Method 1 : - interim stages

Find output from motor ‘A’ kJ/s; (this must be the required input to the gearbox).

A = 20.76/ 0.87 = 23.862 kJ/s which must be the motor output

Motor input = 23.862 kJ/s / 0.75 = 31.82 kJ/s

motor

η = 75%Input 20.76 kJ/sgearbox

η = 87%

‘A’ kJ/s

JR/2008

Work & Energy

Soln.

Method 2 : - single stage

Motor input = 20.76 kJ/s / (0.87 x 0.75) = 31.82 kJ/s

Note: 87 % = 87/100 = 0.87

motor

η = 75%Input 20.76 kJ/sgearbox

η = 87%

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Work & Energy

POTENTIAL ENERGY PE

This is the energy possessed by a body due to its position in space

The amount of energy held will depend upon the height of the body above a specified datum.

Work done = Force x distanceWD = Energy = Weight x heightEnergy = mg x h

Potential Energy (PE) = mghWhen m = mass kg

g = 9.81 ms-2

h = height m

CoG

datum

Height hAbove datum

JR/2008

Work & EnergyKINETIC ENERGY KE

This is the energy possessed by a body due to its motion in space

The amount of energy held will depend upon the mass and the speed of the body.

For this to occur then Newton’s 1st law suggests an opposing force must be applied. The force needed to cause the deceleration = ma

The rate of deceleration = v/t thus Force needed = m v/t

The distance moved in this period = ½ vt

As Work = Force x distance = m v/t x vt/2 = ½ mv2

KINETIC ENERGY = ½ mv2

Consider mass m kg, moving at velocity v ms-1 which is brought to rest in a time of t seconds

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Work & Energy

Examples

1. A crane requires 29.5 kJ of energy to raise a mass of 500 kg to a certain height. Determine the height of the load after the crane has stopped lifting.

Soln.

The crane will do work lifting the mass against gravity (weight) and the

load will gain potential energy mgh

From PE = mgh the height gain h = PE/mg

When PE = 29.5 kJ = 29500 J

m = 500 kg

g = 9.81 ms-2

Height h = 29500/ (500 x 9.81) = 6.01 metres

JR/2008

Work & Energy

Examples

2. A motor car of mass 900 kg is travelling at 50 km/hr

Determine: a. The kinetic energy of the vehicle

b. The speed of the car after it has made a brake application in which 45 kJ of energy is lost

Soln.

a. KE = ½ mv2 when: m = 900 kg

v = 50 x (1000/3600) = 13.889 ms-1

KE = ½ x 900 x 13.8892 = 86806.9 J = 86.81 kJ

b. If the brakes take out 45 kJ of energy the car is left with:-

86.81 kJ - 45 kJ = 41.81 kJ this is in the form of KE ( ½ mv2)

Transpose for v = = = 9.64 ms-1 (34.7 km/hr)

m

KE 2

900

41810 x 2

JR/2008

Work & Energy

Examples

3. A body of mass 800 kg is hauled up an incline plane through a vertical distance of 20m. If the work done against friction is 12 kJ calculate the total work done and the potential energy of the body at its elevated position.

Soln. (tip - work out the second part first)

Work done lifting a height of 20m is equal to the potential energy gain

PE = mgh when:- m = 800 kg

g = 9.81 ms-2

h = 20m

PE = 800 x 9.81 x 20 = 156960 J = 156.96 kJ

BUT 12 kJ of energy was used to overcome the friction on the incline

Thus total energy used to lift = 156.96 + 12 = 168.96 kJ

JR/2008

Work & EnergyExamples

4. A railway train of mass 300 tonnes is travelling at 80 km/hr

when its brakes are applied to reduce its reduce its speed to 35 km/hr

Determine the amount of energy used in the brake application.

Soln.

Initial KE = ½ mv2 when: m = 300 tonnes = 300 x 103 kg

v = 80 x (1000/3600) = 22.222 ms-1

KE = ½ x 300x103 x 22.2222 = 74072592.6 J = 74.07 MJ

Final KE = ½ mv2 when: v = 35 x (1000/3600) = 9.722 ms-1

KE = ½ x 300x103 x 9.7222 = 14177592.6 J = 14.18 MJ

The energy taken out by the brakes is the ‘loss’ in Kinetic Energy

Change in energy (∆ KE) = Initial – Final = 74.07 – 14.18 = 59.89 MJ