Introduction
we apply work in our daily life. In this chapter, we will study work, different types of energy ,such as kinetic energy, potential energy, conservation of energy and power.
3.1 Work
> Definition
Work is something done whenever it transfer energy from one form to another. It also can be defined as force acting on a body moves the body.
> Formulae of work in different situation
(a) W = Fs
(b) W = F cos q s
(c) Work can be calculated by getting the area under the force – displacement graph.
(d) Work due to spring extension (pemanjangan)
W = Fe; F = force; e = spring extension
extension , e = yx
(e) The work of gas inside the container
DW = FDx
= P ADx
= P V
3.2 Energy
Kinetic energy, K

Potential energy, U

Kinetic energy is the energy of an object due to motion.
Kinetic energy stored in an object that is moving equals to the amount of work done accelerating it to that speed.

Potential energy is the energy of an object due to its relative position or physical state.
P.E = mgh
Potential energy stored in objects equals the work done to lift the object against a gravitational field.
Work done = m x g x Dh

– According to principle of conservation energy, energy cannot be created or destroyed, but it can be converted to other form of energy.
K + U = constant
DK = –DU
For example when a ball falls from a building and the ball rebounds, the energy conservation is shown in the diagram below.
Forces can be calculated by differentiating the potential energy.
Example 1
A car is pushed 31 m along a horizontal surface. It the total frictional force is a constant 300N and the work done by the person pushing the car is 14 kJ, how much energy is transform to the kinetic and heat energies ?
solution
the work done against friction is 300 N x 31 m = 9 300 J
this become heat energy.
If total work done is 14 000 J, then 14 000 J – 9300 J = 4700 J must have become kinetic energy.
Example 2
Calculate the retarded force on the object that is shown in the figure below.
Solution
The force is obtained by calculating the slope/gradient (kecerunan) of the graph
Example 3
A force of 10 kN is used to push a body weighing 30 kN up a slope. The distance moved up the
is 40 m. The total height gained is 10 m. If you assume that there is no friction between the slope and the body, how much of the work done ends up as kinetic energy of the body?
Solution
The, total work done by the force pushing the body up the slope is:
Work = force x distance
=10 x 10^{3} x 40m = 400 kJ
The gain in potential energy due to it going up the slope is:
E_{p} = weight x height
= 30 x 10^{3} x 10m = 300 kJ
Total work done that become kinetic energy = 400 kJ – 300 kJ = 100 kJ
Example 4
An object is thrown vertically upwards, what graph will describe its variation of E_{p} and E_{k} with height’?
Example 5
If an object of mass 40 kg is falling at its terminal velocity of 100 ms t through air, what is the energy lost to heat each second as it falls? (g = 9.81 ms^{2})
Solution
The object is losing E_{P} as it falls. When it is at its terminal velocity, the lost E_{P} is all becoming heat. None of it is becoming E_{k} as the object is not gaining E_{k} because it is not speeding up.
Each second the object loses 100 m of height. Hence, E_{P} lost each second is:
E_{P} = mgh
= 40x 9.81 x 100
= 39 240 J
= 39 kJ
Example 6
If an object of mass m is thrown vertically upwards with velocity v and reaches height h, what height would a second object with mass 2 m reach if it is thrown with velocity 4v ?
Solution
The E_{k} becomes E_{P} so: mgh = mv^{2}
cancel off the m and reorganizes to h =
That’s for the first object. z
For the second object: h’ = = = 16 h
In this case, the change in the mass of the object does not matter because the object will have twice the mass but also twice the E_{k} initially as it is going at twice the speed.
Example 7
The graph shows a movement of object 2 kg after it is applied with a force F which changes with displacements. Calculate:
(a) work done by the force.
(b) The height the object could travel.
Solution
Work done by the force = Area under (Fx) graph
= 6 x 20+ (2) x 20 + x2(10) = 130 J
(b) mgh = work done
2 x 9.81 x h = 130 J
h = 6.625 m
3.3 Power
> Definition
Power is the rate of doing work and time taken
Unit: watt (W), joule per second (Js^{1})
Example 8
A wheel is connected to a motor and turned clockwise at a constant speed of 10 revolutions per second. Two weights are hung from a string passing over the wheel as shown. With the wheel turning, the system is in equilibrium. The length of string in contact with the wheel is 0.5 m. What power must the motor produce to turn the wheel?
Solution
Use the equation Power = Force x Velocity.
Here, the resultant force due to the weights would be 20 N down on the left hand side, but the friction between the wheel and the I string must provide enough force to balance the system. i.e. I friction must be 20 N.
If 0.5 m of string is in contact with the wheel and the wheel rotates at 10 revolutions per second, then the wheel rubs on the string for a total distance of 10 s^{1} x 0.5 m = 5 ms^{1} .
So, Power = Force x velocity = 20N x 5ms^{1} =100 W
Example 9
What is the maximum power that you could achieve from a wind turbine that has a blade radius of 14 m in a wind of velocity 2 ms^{1}? (Take the density of air as 1 kgm^{3})
Solution
The energy for the turbine comes from the kinetic energy of the air. The maximum power achievable is if the blade removes 100% of the kinetic energy of the air each second.
In one second, the mass of air passing the blade (which is the air that can give its energy to the blade) is found by calculating the volume of air that passes the blade and multiplying it by the density of the air.
Mass of air = volume of air x density of air
= (pr^{2} x velocity) x density
where r is the radius of the blades.
Substitute in the numbers and get:
Mass of air = 1232 kg per second.
Hence, E_{k} available per second = (1232 kgs^{1}) x 2^{2}
= 2464 W
And finally the power is the energy per second which is 2464 W.
Example 10
Calculate the work done within 20 seconds.
Solution
To find the area under the graph you need to multiply Power x Time (that’s the quantity on one axis times the quantity on the other axis).
you need to recognize that
Work = Power x Time
Hence the units of the quantity that you calculate must be the units of work, i.e. joules.
= 150 J
3.4 Efficiency (kecekapan)
Definition
The efficiency of a machine is the ratio of the useful work done to the energy input.
> Formulae
Work done x 100%
Energy input
Work done per second x 100%
Energy input per second
Output power x 100%
Input power
Useful energy out x 100%
Total energy in
When work done = energy input \ 100% efficiency.
However, most machines do not achieve 100% efficiency because heat is produced and energy is needed to overcome friction.
Example 11
A motor of efficiency 80% is used to haul a lift up its shaft in 10 seconds. It the total length of the shaft is 50 m and the mass of the lift is 1000 kg, what will the electrical power supplied the motor be?
Solution
The weight of the lift is ,W = mg = 1000 kg x 10 N kg^{1} = 10 kN
The power needed to raise the lift is
But that is the output power of the motor. We need to put in more than that as it is only 80% efficient. So the electrical input power is:
Example 12
A machine is used to transport goods at Port Klang. The machine is equipped with 15 A and 30 V. The machine uses 10 s to move goods of 10 kg from ground to the stage which is 40 m above the ground.
Calculate:
(a) Work done by machine.
(b) Potential energy at goods.
(c) Efficiency of machine.
If the goods are transported to cargo at ground with a slop, calculate the velocity before reaching the ground. [g = 10 ms^{2}]
Solution
(a) P = VI
= 30V x 15A
= 450 W
W = P x t
W = 450W x 10s
= 4500 J = 4.5 kJ
(b) Potential energy = mgh
=10kg x10ms^{2} x 40m = 4000 J
Example 13
The graph shows the work done by a machine and the input when lifting a cargo in Port Klang
Calculate:
(a) the efficiency of the machine.
(b) Discuss the efficiency of the machine.
solution
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