This is energy that is usually associated with an electric current and is sometimes referred to incorrectly as
electricity. For example the thermal energy from a chemical
reaction (chemical energy) can be used to boil water and produce steam. The kinetic energy of the molecules of steam (thermal energy) can be used to rotate magnets and this rotation generates an electric current. The electric current transfers the energy to consumers where it is transformed into for example thermal and light energy (filament lamps) and kinetic energy (electric motors).We shall learn later that these different forms of energy all fall into the category of either potential or kinetic energy and are all associated with one or other of the fundamental forces.
Energy can be transformed from one form into another and as far as we know energy can never be created nor can it be destroyed. This is perhaps one of the most fundamental laws of nature and any new theories which might be proposed must always satisfy the principle of energy conservation. A simple example of the principle is, as we have seen, to be found in the transformation of gravitational potential energy into kinetic energy.
2.3.8 ELASTIC
ANDINELASTIC
COLLISIONS
Collisions in which mechanical energy is not lost
are called elastic collisions whereas collision in which mechanical energy is lost are called inelastic collisions. In the real (macroscopic) world mechanical energy is always lost during a collision. However, some collisions
do approximate quite well to being elastic. The collision of two snooker (pool) balls is very nearly elastic, as is the collision between two steel ball bearings. An interesting situation arises when the balls are of the same mass and one is at rest before the collision and the collision takes place along a line joining their centres as shown Figure 251.
u1 = u u2 = 0 v1 = v v2 = V
Before After
m m m m
Figure 251 Rolling balls
Suppose that the speed of the moving ball is u and that the respective speeds of the balls after collision are v and
V. If we now apply the laws of momentum and energy
conservation we have conservation of momentum:
mu = mv + mV
conservation of energy
_ 12 mu 2
= 1_ 2 mv 2
+ 1_ 2 mV 2
From which we see that u = v + V and u2 = v2 + V2
The only solution to these equations is that u = V and v = 0.
This means that the moving ball comes to rest after collision and the ball that was at rest moves off with the speed that the moving ball had before collision. This situation is demonstrated in that well known “toy”, Newton’s Cradle.
2.3.9 Define power.
2.3.10 Define and apply the concept of efficiency.
2.3.11 Solve problems involving momentum, work, energy and power.
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2.3.9 POWER
Suppose that we have two machines A and B that are used in lifting objects. Machine A lifts an object of weight 100 N to a height of 5.0 m in a time of 10 s. When machine B is used to perform the same task it takes 0.1 s. Our instinct
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tells us that machine B is more powerful than machine A. To quantify the concept of power we define power as
Power = the rate of working
i.e., power =power = work _____ time
The unit that is used to measure power is the joule per second which is called the watt, (W) after the 19th Century Scottish engineer James Watt.
In the example above, of our two machines, A will have a power output of 50 W and machine B a power output of 5000 W.
In the following example some sort of engine is used to pull an object at a constant speed along the horizontal.
Engine object
v
Friction
F
Figure 252 Force against friction
The pulling force F (which is the tension in the rope) produced by the engine will be equal to the frictional force between the object and the floor.
Suppose that the engine moves the object a distance Δs in time Δt . The work done against the frictional force (i.e., the work done by the engine) is:
W
Δ = F sΔ
The power P developed by the engine is therefore
P ΔW t Δ --- F Δs t Δ --- ⋅ = = But, v Δs t Δ ---
= hence, we have that
P = Fv
Example
A diesel locomotive is pulling a train at its maximum speed of 60 m s-1. At this speed the power output of the
engine is 3.0 MW. Calculate the tractive force exerted by the wheels on the track.
Solution
Using the formula P = Fv, we have that F P.
v --- = Hence, F 3000000 . 60 --- 50 kN = =
Our answer is at best an approximation since the situation is in fact much more complicated than at first glance. The train reaches a maximum speed because as its speed increases the frictional force due to air resistance also increases. Hence, at its maximum speed all the energy produced by the motors is used to overcome air resistance, energy lost by friction between wheels and track and friction between moving parts of the motors and connected parts.
Again, we shall see that the concept of power and its application is discussed in much more detail in Chapter 8 and elsewhere in the syllabus.
2.3.10 EFFICIENCY
(Machines and efficiencies are discussed in more detail in Chapters 8 and 10).
The efficiency of an engine is defined as follows
Eff = W_____ OUT WIN = POUT ____ P IN
Where WOUT is how much useful work the engine produces and WIN is how much work (energy) is delivered to the engine. The ratio of these two quantities is clearly the same as the ratio power output of the engine to its power output.
To understand the idea of efficiency, we will look at the following example.
Example
An engine with a power output of 1.2 kW drags an object of weight 1000 N at a constant speed up an inclined plane that makes an angle of 30° with the horizontal. A constant frictional force of 300 N acts between the object and the plane and the object is dragged a distance of 8.0 m.
Determine the speed of the object and the efficiency of the engine.
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Solution
We first need to draw a diagram to visualise the situation.
1000 N 1000 × sin30° = 500 N 300 N h = 8.0 × sin30° = 4.0 m 8.0 m P R Engine
The component of the object’s weight down the plane is 1000 × sin30° = 500 N.
The total force against which the machine does work is therefore 500 + 300 = 800 N.
Using P = Fv, we have 1200 = 800 × v so that v = 1.5 m s-1.
The machine lifts the object to a height h, where h = 8.0 × sin30° = 4.0 m.
The useful work that is done by the machine is therefore 1000 × 4.0 = 4000 J.
The actual work that is done by the machine is 800 × 8.0 = 6400 J.
The efficiency of the machine is the useful work done divided by the actual work done which is
. 4000
6400
---×100% = 63%
We can also calculate the energy output per second of the fuel used by the machine. The machine has a useful power output of 1.2 kW and if it is 63% efficient then the fuel must produce energy at the rate of
1.2 ____ 0.63 = 1.9 k J s–1
2.3.11 SOLVEPROBLEMS
INVOLVING,
WORK, ENERGY
AND
POWER
Examples
1. Calculate the momentum of a particle of mass 0.06 kg that has a kinetic energy of 3.2 J.
2. Estimate the minimum take-off power of a grasshopper (cicada).
3. An elastic band of length 2d is attached to a horizontal board as shown in the diagram below.
A margarine tub has some weights attached to the bottom of the inside of the tub such that the total mass of the weights and the tub is M. The tub is placed at the centre of the band and is pulled back until the tub makes an angle θ with the band as shown. The tub is then released such that it is projected down the runway for a distance s before coming to rest. The problem is to deduce an expression for the speed with which the tub leaves the band. The force constant of the elastic band is k.
4. A railway truck, B, of mass 2000 kg is at rest on a horizontal track. Another truck, A, of the same mass moving with a speed of 5.0 m s–1 collides
with the stationary truck and they link up and move off together.
Determine the speed with which the two trucks move off and also the loss of kinetic energy on collision. θ θ M 2d direction of travel of tub original position of elastic band
l θ θ M 2d direction of travel of tub original position of elastic band
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Solutions 1. Use EK = p 2 ___ 2m we have that p =√
_____2mEK = √ ____________ 2 × 0.06 × 3.2 = 0.62 N s 2. Estimates mass of grasshopper = 2.0 × 10-3 kgheight to which it jumps 0.50 m
time it takes to develop take-off power = 200 ms Calculation
energy needed to reach 0.50 m = mgh = 2.0 × 10-3 × 10 × 0.50 = 10-2 J
= 0.05 W
3. In the following example we tie together the ideas of energy transformations and the work done by a non-constant force
We first deduce an expression for the extension of the elastic band.
The stretched length from the geometry of the situation is equal to l = d ____
sinθ
The extension, e, is therefore
e = 2l – 2d = 2d ____
sinθ – 2d = 2d
(
1 ____ sinθ – 1)
The stored elastic energy is therefore
E elas = _ 12 ke 2 = 2 1_ k
(
2d(
1 ____ sinθ – 1)
)
2 = 2k d 2(
1 ____ sinθ – 1)
2If we assume that all the energy is transferred to the tub when the tub is released we have that the kinetic energy of the tub, E k , is such that E k = E elas so that
_ 12 mv 2 = 2k d 2
(
1 ____ sinθ – 1)
2 v 2 = 4k ___ m d 2(
1 ____ sinθ – 1)
2So that the speed is given by v = 2
√
__
k
__ m d
(
1 ____ sinθ – 1)
The above example can form the basis for a useful experiment in which you can investigate the factors that effect the distance that the margarine tub will travel.
4 It helps to start by drawing a diagram:
Before collision
After collision
u
1=
5.0
u
2=
0
V
A
B
A
B
As the trucks couple after the collision, the conservation of momentum law states that:
m1u1+m2u2 = (m1+m2)V , , e r e h w m1 = m2 = 2000 so that 2000×5+0 = (2000+2000)V = 2.5 m s–1. V ⇒ 10000 4000 --- =
The total KE before collision equals J 1
2
---×2000 ×52+0 = 25000
The total KE after collision equals 1
2
---×4000×(2.5)2 = 12500
So that the kinetic energy lost on collision is (25000 – 12500) = 12500 J.
By lost energy we mean that the energy has been dissipated to the surroundings. Some of it will be converted into sound and most will heat up the coupling between the trucks.
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Exercise 2.3
1. Suppose that in Example 4, (page 64) after the collision, truck A and truck B do not link but instead truck A moves with a speed of 1 m s–1 and
in the same direction as prior to the collision.
Determine
(i) the speed of truck B after collision (ii) the kinetic energy lost on collision
2. A man drags a sack of flour of mass 100 kg at constant speed up an inclined plane to a height of 6.0 m. The plane makes an angle of 30°with the horizontal and a constant frictional force of 250 N acts on the sack down the plane.
Determine the efficiency of the inclined plane?
3. A car of mass 1000 kg is parked on a level road with its handbrake on. Another car of mass 1500 kg travelling at 10 m s–1 collides into the back
of the stationary car. The two cars move together after collision in the same straight line. They travel 25m before finally coming to rest.
Determine the average frictional force exerted on the cars as they come to rest.
2.4 UNIFORM
CIRCULAR
MOTION
(This topic links with Topics 6.3 and 9.4)
2.4.1 Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed toward the centre of the circle
2.4.2 Apply the expression for centripetal acceleration.
2.4.3 Identify the force producing circular motion in various situations.
Examples include gravitational force acting on the Moon and friction acting sideways on the tyres of a car turning a corner.
2.4.4 Solve problems involving circular motion.
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Problems on banked motion (aircraft and vehicles going round banked tracks) will not be included.
2.4.1 CIRCULARMOTIONANDVECTOR
DIAGRAMS
We have mentioned earlier in this chapter that when a car goes round a circular bend in the road moving with constant speed it must be accelerating. Acceleration is a vector and is defined as the rate of change of velocity. Velocity is also a vector and so it follows that if the direction of motion of an object is changing then even though its speed might be constant, its velocity is changing and hence it is accelerating. A force is needed to accelerate an object and we might ask what is the origin of the force that causes the car to go round the bend in the road. Do not fall into the trap of thinking that this force comes from the engine. This would imply that if you are on a bicycle then you can only go round a corner if you keep pedalling. To understand the origin of the force just think what happens if the road is icy. Yes, it is this friction between the tyres and the road that produces the force. If there is no friction then you cannot negotiate a bend in the road.
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Let us think of the example where you whirl an object tied to a string about your head with constant speed. Clearly the force that produces the circular motion in this case in a horizontal plane is the tension in the string. If the string were to snap then the object would fly off at a tangent to the circle. This is the direction of the velocity vector of the object. The tension in the string acts at right angles to this vector and this is the prerequisite for an object to move in a circle with constant speed. If a force acts at right angles to the direction of motion of an object then there is no component of force in the direction of motion and therefore no acceleration in the direction of motion. If the force is constant then the direction of the path that the object follows will change by equal amounts in equal time intervals hence the overall path of motion must be a circle. Figure 256 shows the relation between the direction of the velocity vector and the force acting on a particle P moving with constant speed in a circular path.
F P v O P v O Q r θ Δ
Figure 256 Centripetal Force
The force causing the circular motion is called the centripetal
force and this force causes the particle to accelerate
towards the centre of the circle and this acceleration is called the centripetal acceleration. However, be careful to realise that the centripetal acceleration is always at right angles to the velocity of the particle. If the speed of the particle is reduced then it will spiral towards the centre of the circle, accelerating rapidly as it does so. This is in effect what happens as an orbiting satellite encounters the Earth’s atmosphere.
People sometimes talk of a centrifugal force in connection with circular motion. They say something along the lines that when a car goes round a bend in the road you feel a force throwing you outwards and this force is the centrifugal force. But there is no such force. All that is happening is that you are moving in accordance with Newton’s laws. Before the car entered the bend you were moving in a straight line and you still want to keep moving in a straight line. Fortunately the force exerted on you by a side of the car as you push up against it stops you moving in a straight line. Take the side away and you will continue moving in a straight line as the car turns the bend.