2.3.1 Outline what is meant by work.
2.3.2 Determine the work done by a non- constant force by interpreting a force–
displacement graph. A typical example
would be calculating the work done in extending a spring. See 2.3.7
2.3.3 Solve problems involving the work done by a force.
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2.3.1 OUTLINEWHATISMEANTBYWORK
The terms work and energy are used frequently in everyday speech. However, although it is possible to give an exact physical definition of ‘work’ it is not so easy to define precisely what is meant by ‘energy. We often use the term in the context of ‘having enough energy to get something done’. So for the time being, let us be happy with this
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idea and also note that getting ‘something done’ usually involves a transfer of energy from one form to another. For example a car engine uses fuel in order to get the car moving and to keep it moving, you eat food in order to live, oil is burned in a power station to provide electrical energy and you can think of other examples.
h E ngine mg 2 mg 2h 2 mg mg
Figure 244 Fuel used to lift a weight
In Figure 244 we imagine a situation in which we are using fuel in an engine whose function is to lift a weight mg to a certain height.
In the first diagram the engine lifts the weight to a height
h and in the second diagram the engine lifts a weight 2mg
to a height 2h. In the first situation let us suppose that the engine requires 1 dm3 of fuel to complete the task. Then in
the second situation we would guess that the engine would require 4 dm3 of fuel. The amount of fuel used is a measure
of the energy that is transferred and to complete the tasks we say that the engine has done work. The amount of work that is done will be a measure of the energy that has been transferred in the performance of the tasks. Clearly the engine has done work against a force, in this situation, the force of gravity. If we double the force i.e. the weight, then the work done is doubled and if we double the distance through which the weight is moved then the work done is also doubled. If we double both together, then the work done is quadrupled. It seems that we can define work as:
Work = force × distance moved
However, we have to be careful in using this definition. Consider the situation below in which the engine lifts the weight up a slope to height h.
s m g
Engine
h
θ
Figure 245 Using a slope
We shall assume that the surface of the slope is frictionless. The work W done by the engine using the above definition is:
W = force × s
where s is the distance up the slope.
The force this time is the component of the weight down the slope.
Hence, W = mg sin θ × s
But sin θ = h __ s ⇒s = ____ h sin θ
meaning that W = mg sin θ × s = mg sin θ × h ____ sin θ = mgh
That is,
W = mgh
This is just the amount of work that the engine would have performed if it had lifted the weight directly. Our definition of work therefore becomes
Work = magnitude of the force × displacement in the direction of the force
Which is the same as
work = magnitude of the component of the force in the direction moved × the distance moved
The SI unit of work is the newton metre and is called the joule named after the 19th Century physicist James
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2.3.2 THE
WORK
DONE
BY
A
NON-
CONSTANT
FORCE
Clearly, when a spring is stretched or compressed, work is done. That a stretched spring or a compressed spring has the potential to do work is apparent from many everyday examples and it is left as an exercise for you to describe some of the examples.
Figure 246 (a) below shows a plot of a constant force against displacement and Figure 246 (b) shows a plot of the spring force against displacement.
Graph 1 Graph 2 x F orc e F F orce x d s F
Figure 246 (a) and (b) Force vs displacement graphs
The area under each of the graphs is clearly equal to the work done. In Figure 256 (a) when the force F undergoes a displacement d the work done is Fd. In Figure 246 (b) when the force F produces an extension s then the work done is
1 2
---Fs. But in this case F = ks, hence the work done is
1 2 ---×( ) sks × 1 2 ---ks2 =
The work that has been done is stored in the spring as elastic potential energy Eelas . The adjective potential in this context essentially means “hidden”. Clearly
Eelas = 1 __ 2 ks2
We can extend this idea to find the work done by any non-constant force. If we know how the force depends on displacement then to find the work done by the force we just compute the area under the force-displacement graph.
2.3.3 SOLVEPROBLEMS
INVOLVING
THEWORK
DONE
BY
A
FORCE
Example
A force of 100 N pulls a box of weight 200 N along a smooth horizontal surface as shown in the Figure below.
45° 100 N
Calculate the work done by the force
(a) in moving the box a distance of 25 m along the horizontal
(b) against gravity.
Solution
(a) The component of the force along the direction of motion, i.e., the horizontal component, Fh, can be determined by using the fact that
cos 45˚ = ___ F h
100
⇔ F h = 100 cos45˚
= 71 N
That is, the component of the force along the direction of motion is 71N. Therefore the work done (F × s) = 71 × 25 = 1780 That is, the work done is 1780 J.
(b) There is no displacement by the force in the direction of gravity. Hence the work done by the force against gravity is zero.
Suppose that a constant frictional force of 50 N acts on the box. How much work is done against friction? Again, using the fact that W = force × distance, we have that the work done is simply 50 × 25 = 1250 N.
Notice that in the above example we use the expressions ‘work done by’ and ‘work done against’. This occurs over and over again in physics. For example the engine lifting the weight does work against gravity. However, if
45° 100 N
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the weight is allowed to fall, the work is done by gravity. Strictly speaking we have a sign convention for work. The convention is that
Work done on a system is negative.
Work done by a system is positive.
In the example above, the box can be identified as the system.
Although this convention is not always important in many mechanics situations, it is of great importance when we come to consider the relationship between heat and work (Chapter 10) and in the relation between field strength and potential (Chapter 9)
Exercise 2.2 (g)
A man drags a sack of flour of mass 50 kg at constant speed up an inclined plane to a height of 5.0 m. The plane makes an angle of 30° with the horizontal and a constant frictional force of 200 N acts on the sack down the plane.
Calculate the work that the man does against:
(a) friction. (b) gravity.
2.3.4 Outline what is meant by kinetic energy.
2.3.5 Outline what is meant by change in gravitational potential energy.
2.3.6 State the principle of conservation of energy.
2.3.7 List different forms of energy and describe examples of the transformation of energy from one form to another.
2.3.8 Distinguish between elastic and inelastic collisions.
© IBO 2007
2.3.4 KINETICENERGY
A moving object possesses the capacity to do work. A hammer, for example, by virtue of its motion can be used to do work in driving a nail into a piece of wood.
So how do we find out just how much work a moving object is capable of doing?
In the diagram below, a force F moves an object of mass m a distance d along a horizontal surface. There is no friction between the object and the surface.
d
F
m
Figure 249 An example of kinetic energy
We will assume that the object starts from rest. The force
F will accelerate the object in accordance with Newton’s
Second Law and the magnitude of the acceleration will be given by using the formula
F = ma, from which we obtain a = F __ m .
We can use the equation v2 = u2 + 2as to find an expression
for the speed of the object after it has moved a distance
d (= s).
That is, v2 = 02 + 2
(
F __m
)
d ⇒ v2 =2 Fd ___ mFrom which Fd = 1 __ 2 mv2
The work done (Fd) by the force F in moving the object a distance d in the direction of F is now, therefore, expressed in terms of the properties of the body and its motion.
The quantity 1 __ 2 mv2 is called the kinetic energy (KE) of the
body.
This is the energy that a body possesses by virtue of its motion. The kinetic energy of a body essentially tells us how much work the body is capable of doing.
We denote the kinetic energy by EK, so that EK = 1 __ 2 mv2 A very useful relationship exists between kinetic energy and momentum.
We have that p = mv such that p2 = m2v2
Hence substituting for v2, we have
EK = p 2 ___ 2m and p2 __ m = mv2 = 2 EK
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2.3.5 POTENTIALENERGY
When an object is resting on top of a wall it has the potential to do work. If it falls off the wall onto a nail sticking out of a piece of wood then it could drive the nail further into the wood. Work is needed to lift the object on to the top of the wall and we can think of this work as being ‘stored as
potential energy in the object.
In general if an object of mass m is lifted to a height h above the surface of the earth then the work done is mgh and the object has a potential energy also equal to mgh (force = mg, displacement = h). However, we have to be careful when using this equation since if h is too large then we can no longer regard g as being constant. Also, we must note that there is no zero of potential energy in this situation. (In chapter 9 we will see how we define the zero of potential energy). If the object rests upon the ground and we dig a hole, then the object has potential to do work when it falls down the hole. The object in fact has gravitational potential energy by virtue of its position in the Earth’s gravitational field. No matter where an object is placed in the Universe it will be attracted by the gravitational force of the Earth. We say that the Earth has an associated gravitational field. (Indeed all objects will have an associated gravitational field as we will discuss in the chapters 6 and 9). At this stage you should appreciate that when an object is moved a distance Δh in the Earth’s gravitational field and in the direction of the field, its change in potential energy ΔEP is:
ΔEP = mgΔh
(provided that g is constant over the distance moved.)
Another example of potential energy is, as we saw in the Section 2.3.2, elastic potential energy. To summarise therefore we have: ΔEP = mgΔh and Eelas = 1 __ 2 ks2
2.3.6 THE
PRINCIPLEOFENERGY
CONSERVATION
(In Topic 7.3 we shall see that this should actually be the principle of mass-energy conservation)
To demonstrate the so-called principle of energy
conservation we will solve a dynamics problem in
two different ways, one using the principle of energy conservation and the other using Newton’s laws and the kinematics equations. Example
Example
An object of mass 4.0 kg slides from rest without friction down an inclined plane. The plane makes an angle of 30° with the horizontal and the object starts from a vertical height of 0.50 m. Determine the speed of the object when it reaches the bottom of the plane.
Solution
The set-up is shown in this diagram:
We set ‘zero–level’ at point B, the base of the incline, so that h = 0 and so that at point A, h = 0.50. At A, the object has no kinetic energy (v = 0) and at point B, the object has gained kinetic energy (v = V).
Method 1: Kinematic solution
The force down the plane is given by mgsinθ = 20 N.
Using Newton’s 2nd law (F = ma) gives the acceleration
a = 20 ___
4.0 = 5.0 m s–2
(Note that we could have determined the acceleration by writing down the component of g down the plane)
Using V 2
= u 2
+ 2as with u = 0, and s (the distance down
the plane) s = 0.50 ____ sin θ = 0.50 ______ sin 30˚ = 1.0 We have, V 2 = 0 2 + 2 × 5 × 1 ⇒ V = √ ___ 10 = 3.2 m s-1 30° mg Reaction force 0.50 m h = 0.50 h = 0 v = 0 v = V A B
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Method 2: Energy Principle
As the object slides down the plane its potential energy becomes transformed into kinetic energy. If we assume that no energy is ‘lost’ we can write
change in PE = mgh = gain in KE = _ 12 mv 2
So that 1_ 2 mv 2 = mgh ⇒ V 2 = 2gh
⇒ V =
√
____2ghUsing the values of g and h, we have that V = 3.2 m s-1.
That is, the object reaches a speed of 3.2 m s-1.
Note that the mass of the object does not come into the question, nor does the distance travelled down the plane. When using the energy principle we are only concerned with the initial and final conditions and not with what happens in between. If you go on to study physics in more depth you will find that this fact is of enormous importance.
The second solution involves making the assumption that potential energy is transformed into kinetic energy and that no energy is lost. This is the so–called ‘energy principle’, this means the energy is conserved.
Clearly in this example it is much quicker to use the energy principle. This is often the case with many problems and in fact with some problems the solution can only be achieved using energy considerations.
What happens if friction acts in the above example? Suppose a constant force of 16 N acts on the object as it slides down the plane. Now, even using the energy principle, we need the distance down the plane so we can calculate the work done against friction
This is 16 × 1.0 = 16 J. The work done by gravity i.e. the change in PE = 20 J
The total work done on the object is therefore
20 – 16 = 4.0 J.
Hence the speed is now given by _ 12 mv 2 = 4.0. To give v = 1.4 m s-1.
So in this problem, not all the work done has gone into accelerating the object. We say that the frictional force has
dissipated energy. If we are to retain to the idea of energy
conservation then we must “account” for this “lost” energy.
It was the great triumph of some late eighteenth and early nineteenth physicists and engineers to recognise that this ‘lost energy’ is transformed into thermal energy. If you rub your finger along the top of a table you will definitely feel it getting warm. This is where the “lost” energy has gone. It has in fact been used to make the molecules of the table and the molecules of your finger vibrate more vigorously.
Another thing to notice is that this energy is “lost” in the sense that we can’t get it back to do useful work. If there were no friction between the object and the surface of the plane then, when it reached the bottom of the plane, work could be done to take it up to the top of the plane and this cycle could go on indefinitely. (We could actually set up the arrangement such that the objects KE at the bottom of the plane could be used to get it back to the same height again). It is what we call a reversible process. The presence of friction stops this. If the object is dragged back up the plane you won’t get the energy back that has been “lost” due to friction, you will just “lose” more energy. This is an
irreversible process. We can now start to glimpse why, even
though it is impossible to destroy energy, it is possible to “run out” of “useful energy”. Energy becomes as we say,
degraded.
The general principle of energy conservation finds its formulation in the ‘First Law of Thermodynamics’ and the consequences of this law and the idea of energy degradation and its implication on World energy sources is discussed in much more detail in Chapter 8.
2.3.7 TYPESOFENERGYANDENERGY
TRANSFORMATIONS
(This links with Topic 8)There are many different forms of energy and their transformations of which some examples are given here.