At the end of this section you should be able to: a. b. c. d. e. f. describe potential energy as energy due to position and derive potential energy as mgh describe kinetic energy as energy due to motion and derive kinetic energy as mv2/2 state conservation of energy laws and solve problems where energy is conserved define power as rate of energy transfer define couple, torque and calculate work done by variable force or torque solve problems where energy is lost due to friction

Table of contents: 3 3.1 3.2 3.3 3.3.1 3.3.2 3.3.3 3.3.4 3.3.5 3.4 3.5 3.6 3.6.1 3.6.2 3.6.3 Work, Power and Energy......................................................................................................................................1 Work done by a constant force .............................................................................................................................2 Work done by a variable force..............................................................................................................................3 Energy ..................................................................................................................................................................3 Potential Energy..........................................................................................................................................4 Formulae for gravitational potential energy................................................................................................4 Kinetic energy.............................................................................................................................................5 Formulae for kinetic energy........................................................................................................................5 Kinetic energy and work done ....................................................................................................................6 Conservation of energy.........................................................................................................................................7 Power....................................................................................................................................................................8 Moment, couple and torque ................................................................................................................................11 Work done by a constant torque................................................................................................................12 Power transmitted by a constant torque ....................................................................................................13 Work done by a variable torque ................................................................................................................15

Page numbers on the same topic in, Applied Mechanics, 3rd Edition, Hannah & Hillier Secti notes All of Section 3 Section in Hannah & Hillier Chapter 10 Excluding sections 10.4, 10.9, 10.11, 10.16, 1.17 Page No. in Hanna & Hiller 180 - 210

Fundamentals of Mechanics – Kinetics: Section 3 - Work, Powers and Energy

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3.1

Work done by a constant force

When the point at which a force acts moves, the force is said to have done work. When the force is constant, the work done is defined as the product of the force and distance moved.

work done = force × distance moved in direction of force

Consider the example in Figure 3.1, a force F acting at the angle θ moves a body from point A to point B.

F sθ s co

A

θ s

B

Figure 3.1: Notation for work done by a force The distance moved in the direction of the force is given by

Distance in direction of force = s cosθ

So the work done by the force F is

Work done = F s cosθ

Equation 3.1 If the body moves in the same direction as the force the angle is 0.0 so Work done = Fs When the angle is 90 then the work done is zero. The SI units for work are Joules J (with force, F, in Newton's N and distance, s, in metres m). Worked Example 3.1 How much work is done when a force of 5 kN moves its point of application 600mm in the directio Solution the force.

work done = 5 × 10 3 × 600 × 10 −3 = 3000 J = 3 kJ

Worked Example 3.2

(

)(

)

Find the work done in raising 100 kg of water through a vertical distance of 3m. Solution The force is the weight of the water, so

Fundamentals of Mechanics – Kinetics: Section 3 - Work, Powers and Energy

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work done = (100 × 9.81)× 3 = 2943 J

3.2 Work done by a variable force

Forces in practice will often vary. In these cases Equation 3.1 cannot be used. Consider the case where the force varies as in Figure 3.2 For the thin strip with width ds - shown shaded in Figure 3.2 - the force can be considered constant at F. The work done over the distance ds is then

work done = F ds

This is the area of the shaded strip. The total work done for distance s is the sum of the areas of all such strips. This is the same as the area under the Forcedistance curve.

force

F

0

ds

s

distance

Figure 3.2: Work done by a variable force So for a variable force

work done = area under force/distance curve

Equation 3.2

Clearly this also works for a constant force - the curve is then a horizontal line. In general you must uses some special integration technique to obtain the area under a curve. Three common techniques are the trapezoidal, mid-ordinate and Simpson's rule. They are not detailed here but may be found in many mathematical text book.

3.3

Energy

A body which has the capacity to do work is said to possess energy. For example , water in a reservoir is said to possesses energy as it could be used to drive a turbine lower down the valley. There are many forms of energy e.g trical, chemical heat, nuclear, mechanical etc. The SI units are the same as those for work, Joules J. In this module only purely mechanical energy will be considered. This may be of two kinds, potential and kinetic.

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3.3.1

Potential Energy

There are different forms of potential energy two examples are: i) a pile driver raised ready to fall on to its target possesses gravitational potential energy while (ii) a coiled spring which is compressed possesses an internal potential energy. Only gravitational potential energy will be considered here. It may be described as energy due to position relative to a standard position (normally chosen to be he earth's surface.) The potential energy of a body may be defined as the amount of work it would do if it were to move from the its current position to the standard position.

3.3.2

Formulae for gravitational potential energy

A body is at rest on the earth's surface. It is then raised a vertical distance h above the surface. The work required to do this is the force required times the distance h. Since the force required is it's weight, and weight, W = mg, then the work required is mgh. The body now possesses this amount of energy - stored as potential energy - it has the capacity to do this amount of work, and would do so if allowed to fall to earth. Potential energy is thus given b

potential energy = mgh

Equation 3.3 where h is the height above the earth's surface. Worked example 3.3 What is the potential energy of a 10kg mass: a) 100m above the surface of the earth b) at the bottom of a vertical mine shaft 1000m deep.

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Solution a)

potential energy = mgh = 10 × 9.81× 100 = 9810 J = 9.81kJ potential energy = mgh = 10 × 9.81× (− 1000) = −98100 J = −98.1kJ

b)

3.3.3

Kinetic energy

Kinetic energy may be described as energy due to motion. The kinetic energy of a body may be defined as the amount of work it can do before being brought to rest. For example when a hammer is used to knock in a nail, work is done on the nail by the hammer and hence the hammer must have possessed energy. Only linear motion will be considered here.

3.3.4

Formulae for kinetic energy

Let a body of mass m moving with speed v be brought to rest with uniform deceleration by a constant force F over a distance s. Using Equation 1.4

v 2 = u 2 + 2as 0 = u 2 + 2as s=

And work done is given by

v2 2a

work done = force × distance = Fs =F v2 2a

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The force is F = ma so

v2 work done = ma 2a 1 = mv 2 2

Thus the kinetic energy is given by

kinetic energy =

1 2 mv 2

Equation 3.4

3.3.5

Kinetic energy and work done

When a body with mass m has its speed increased from u to v in a distance s by a constant force F which produces an acceleration a, then from Equation 1.3 we know

v 2 = u 2 + 2as 1 2 1 2 v − u = as 2 2 multiplying this by m give an expression of the increase in kinetic energy (the difference in kinetic energy at the end and the start)

1 2 1 mv − mu 2 = mas 2 2

Thus since F = ma

increase in kinetic energy = Fs

Fs = work done

but also we know

So the relationship between kinetic energy can be summed up as Work done by forces acting on a body = change of kinetic energy in the body Equation 3.5 This is sometimes known as the work-energy theorem.

Worked examp

.4

A car of mass 1000 kg travelling at 30m/s has its speed reduced to 10m/s by a constant breaking force over a distance of 75m. Find: a) The cars initial kinetic energy b) The final kinetic energy c) The breaking force

Solution a)

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Initial kinetic energy =

1 2 mv 2 = 500 × 30 2 = 450000 J = 450 kJ

b)

Final kinetic energy =

1 2 mv 2 = 500 × 10 2 = 50000 J = 50 kJ

c) Change in kinetic energy = 400 kJ By Equation 3.5 work done = change in kinetic energy so

Fs = change in kinetic energy Breaking force × 75 = 400 000 Breaking force = 5333 N

3.4 Conservation of energy

The principle of conservation of energy state that the total energy of a system remains constant. Energy cannot be created or destroyed but may be converted from one form to another. Take the case of a crate on a slope. Initially it is at rest, all its energy is potential energy. As it accelerates, some of it potential energy is converted into kinetic energy and some used to overcome friction. This energy used to overcome friction is not lost but converted into heat. At the bottom of the slope the energy will be purely kinetic (assuming the datum for potential energy is the bottom of the slope.) If we consider a body falling freely in air, neglecting air resistance, then mechanical energy is conserved, as potential energy is lost and equal amount of kinetic energy is gained as speed increases. If the motion involves friction or collisions then the principle of conservation of energy is true, but conservation of mechanical energy is not applicable as some energy is converted to heat and perhaps sound. Worked Example 3.5 A cyclist and his bicycle has a mass of 80 kg. After 100m he reaches the top of a hill, with slope 1 in 20 measured along the slope, at a speed of 2 m/s. He then free wheels the 100m to the bottom of the hill where his speed has increased 9m/s. How much energy has he lost on the hill? Solution

100m h

Figure 3.3: Dimensions of the hill in worked example 3.5 If the hill is 100m long then the height is:

Fundamentals of Mechanics – Kinetics: Section 3 - Work, Powers and Energy

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h = 100

So potential energy lost is

1 = 5m 20

mgh = 80 × 9.81× 5 = 3924 J

1 2 1 1 mv − mu 2 = m v 2 − u 2 2 2 2 = 40(81 − 4) = 3080 J

Increase in kinetic energy is

(

)

By the principle of conservation of energy

Initial energy = Final energy + loss of energy(due to friction etc.) loss of energy (due to friction etc.) = 3924 − 3080 = 844 J

3.5 Power

Power is the rate at which work is done, or the rate at which energy is used transferred.

Power =

work done time taken

Equation 3.6

The SI unit for power is the watt W. A power of 1W means that work is being done at the rate of 1J/s. Larger units for power are the kilowatt kW (1kW = 1000 W = 103 W) and the megawatt MW (1 MW = 1000000 W = 106 W). If work is being done by a machine moving at speed v against a constant force, or resistance, F, then since work doe is force times distance, work done per second is , which is the same as power.

Power = Fv

Equation 3.7

Worked Example 3.6 A constant force of 2kN pulls a crate along a level floor a distance of 10 m in 50s. What is the power used? Solution

Work done = force × distance = 2000 × 10 = 20 000 J

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Power =

work done time taken 20 000 = 50 = 400W

Alternatively we could have calculated the speed first

v=

distance time 10 = = 0.2 m / s 50

and then calculated power

Power = Force × Speed = Fv = 2000 × 0.2 = 400W

Worked Example 3.7 A hoist operated by an electric motor ha of 500 kg. It raises a load of 300 kg vertically at a steady speed of 0.2 m/s. Frictional resistance can be taken to be constant at 1200 N. What is the power required? Solution

Total mass = m = 800 kg Weight = 800 × 9.81 = 7848 N

Total force = 7848 + 1200 = 9048 N

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From Equation 3.7

Power = force × speed = 9048 × 0.2 = 1810W = 1.81kW

Worked example 3.8 A car of mass 900 kg has an engine with power output of 42 kW. It can achieve a maximum speed of 120 km/h along the level. a) What is the resistance to motion? b) If the maximum power and the resistance remained the same what would be the maximum speed the car could achieve up an incline of 1 in 40 along the slope? Solution

Wsinθ N

1200 N

40 1 θ W = 900g

Figure 3.4: Forces on the car on a slope in Worked Example 3.8

Wcosθ

First get the information into the correct units:

120 km / h =

120 × 1000 3600 = 33.33 m / s

a) Calculate the resistance

Power = Force × speed = Resistance × speed 42000 = Resistance × 33.33 42000 Resistance = = 1260 N 33.33

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b)

Total force down the incline = frictional force + component of weight down incline = 1260 + mg sin θ = 1260 + 900 × 9.81 = 1260 + 221 = 1481 N

Power = force × speed Speed = Power force 42000 = 1481 = 28.4 m / s

3600 1000 = 102 km / h

1 40

Or in km/h

Speed = 28.4

3.6

Moment, couple and torque

The moment of a force F about a point is its turning effect about the point. It is quantified as the product of the force and the perpendicular distance from the point to the line of action of the force.

F

d O

Figure 3.4: Moment of a force In Figure 3.5 the moment of F about point O is

moment = Fd

Equation 3.8 A couple is a pair of equal and parallel but opposite forces as shown in Figure 3.6:

Fundamentals of Mechanics – Kinetics: Section 3 - Work, Powers and Energy

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F

p

F

Figure 3.6: A couple The moment of a couple about any point in its plane is the product of one force and the perpendicular distance between them:

Moment of couple = Fp

Equation 3.9 Example of a couple include turning on/off a tap, or winding a clock. The SI units for a moment or a couple are Newton metres, Nm. In engineering the moment of a force or couple is know as torque. A spanner tightening a nut is said to exert a torque on the nut, similarly a belt turning a pulley exerts a torque on the pulley.

3.6.1

Work done by a constant torque

Let a force F turn a light rod OA with length r through an angle of θ to position OB, as shown in Figure 3.7.

F B F s r A θ O

Figure 3.7: Work done by a constant torque The tor TQ exerted about O is force times perpendicular distance from O.

r

TQ = Fr

Equation 3.10

Now work done by F is

s is the arc of the circle, when θ is measure in radians

work done = Fs s = rθ

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work done = Frθ work done = TQθ

Equation 3.11

The work done by a constant torque TQ is thus the product of the torque and the angle through which it turns (where the angle is measured in radians.) As the SI units for work is Joules, TQ must be in Nm

3.6.2

Power transmitted by a constant torque

Power is rate of doing work. It the rod in Figure 3.7 rotates at n revolutions per second, then in one second the angle turned through is

θ = 2πn

radians, and the work done per second will be, by Equation 3.11

work done per second = power = TQ 2πn as angular speed is

ω = 2πn power = 2πnTQ power = ωTQ

Equation 3.12

then

The units of power are Watts, W, with n in rev/s, ω in rad/s and TQ in Nm.

Worked Example 3.9 A spanner that is used to tighten a nut is 300mm long. The force exerted on the end of a spanner is 100 N. a) What is the torque exerted on the nut? b) What is the work done when the nut turns through 30°?

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Solution a) Calculate the torque by Equation 3.10

TQ = Fr

= 100 × 300 ×10 −3 = 30 Nm

b) Calculate the work done by Equation 3.11

(

)

work done = TQθ

π 6 = 15.7 J

= 30 ×

Worked Example 3.10 An electric motor is rated at 400 W. If its efficiency is 80%, find the maximum torque which it can exert when running at 285 Solution Calculate the speed in rev/s using Equation 3.12

power = 2πnTQ n= n=

Calculate the power as the motor is 80% efficient

power 2πTQ 2850 = 47.5 rev / s 60

power = 400 ×

80 = 320W 100

power = 2πnTQ TQ = power 2πn 320 = 2π 47.5 = 1.07 Nm

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