1. ………
2. ………
stationary object
An object moving with uniform velocity
It is in a stationary state
It is moving with uniform velocity
Normal reaction, R Normal reaction, R
Weight, W=mg weight, W=mg
Magnitude of R = W Magnitude of R = mg cos R and W acts in opposite direction. And acts in opposite direction.
So, Resultant force = W – R = 0 So,Resultant force = mg cos – R = 0 ( object in equilibrium ) ( object in equilibrium )
normal reaction, R
friction force force, F
Weight, W Force , F = Frictional force
Resultant force = F – Frictional force
= 0 (object in equilibrium)
m = 200 g s = 3.0 m t = 0.79 s u = 0 g = ? = 0.2 kg
s = ut + ½ g t2
3.0 = 0 (0.7) + ½ g (0.792) g = 9.6 m s-2
The answer less than the constant because of the air frictional force.
23 Addition of Force
1. Addition of force is defined as ...………..
………
………
Examples : the forces are acting in one direction F1 = 10 N F2 = 5 N
Resultant force, F
Example : the forces are acting in opposite directions F1 = 10 N F2 = 5 N
Resultant force, F
Example : the forces are acting in different directions
F2 = 5 N
500 F
F1 = 10 N
Parallelogram method:
1. Draw to scale.
2. Draw the line parallel with F1 to the edge of F2, and the line parallel with F2 to the edge of F1
3. Connect the diagonal of the parallelogram starting from the initial point.
4. Measure the length of the diagonal from the initial point as the value of the resultant force.
a resultant force is a single force the represents in magnitude and direction two or more forces acting on an object F resultant = the total of forces (including the directions of the forces)
= F1 + F2 = 10 + 5 = 15 N
= F1 - F2 = 10 - 5 = 5 N
24 F2
F
F1
Triangle method 1. Draw to scale.
2. Displace one of the forces to the edge of another force.
3. Complete the triangle and measure the resultant force from the initial point.
Example 1: During Sport Day two teams in tug of war competition pull with forces of 6000 N and 5300 N respectively. What is the value of the resultant force?
Are the two team in equilibrium?
Example 2: A boat in a river is pulled horizontally by two workmen. Workmen A pulls with a force of 200 N while workmen while workmen B pulls with a force of 300 N. The ropes used make an angle 250 with each other. Draw a parallelogram and label the resultant force using scale of 1 cm : 50 N.
Determine the magnitude of resultant force.
Solution : Resultant force, F = 6000 – 5300 =700 N
They mere not in equilibrium
Resultant force, F = 10.5 x 50
= 525 N
250
10.5 cm
25 Resolution of a force
1. Resolution of a force is ………
Refer to trigonometric formula:
Example : The figure below shows Ali mopping the floor with a force 50 N at an angle of 600 to the floor.
F = 50 N
Example of resolution and combination of forces
reverse process of finding the resultant force
Fy F is the resultant force of Fx and Fy Therefore, F can be resolved
into Fx and Fy F
Vertical Component
Fx horizontal component
Cos = F Fx
, therefore Fx = F cos Sin =
F Fy
, therefore Fy = F sin
Fx = F cos = 50 cos 60 Fx = 50 (0.5)
= 25 N
Fy = F Sin = 50 sin 600 Fy = 50 (0.8660)
= 43.3 N
F = mg sin 400 + 200 = 800(0.6427) + 200 = 514.2 + 200
= 714.2 N
mg = 800 N 600
F = ?
200 N 400
400
26 Problem solving
1. When a system is in equilibrium, ……….
2. If all forces acting at one point are resolved into horizontal and vertical
components, ………
3. Example 1; Show on a figure;
a) the direction of tension force, T of string b) the resultant force act to lamp
c) calculate the magnitude of tension force, T a)
mlamp = 1.5 kg Wlamp = 14.7 N
Exercise 2.9
1. Two force with magnitude 18 N and 6 N act along a straight line. With the aid of
diagrams, determine the maximun possible value and the minimum possible value of the resultant force.
2. A football is kicked simultaneously by two players with force 220 N and 200 N respectively, as shown in Figure 2.9. Calculate the magnitude of the resultant force.
220 N 900
200 N
T b) T’ T
700 700
the resultant force is equal to zero.
the sum of each component is equal to zero.
(c ) T’ = 2T sin 700 Therefore, mlampg = 2T sin 700
T =
0 lamp
2sin70 g m
=
2sin700
1.5(9.8)
= 7.82 N
= mLamp g
Fmaximum when both of forces act in same direction;
Fmaximum = 18 + 6 18 N 24 N = 24 N 6 N
Fminimum when the forces act in opposite direction ;
Fminimum = 18 – 6 18 N 12 N = 12 N 6 N
F = Resultant of Force F2 = 2202 + 2002 F = 297.32 N
F
27
2.10 UNDERSTANDING WORK, ENERGY AND EFFICIENCY
Work
1. Work is done, ………..
………
2. WORK is the product.……….
………
3. The formulae of work;
4. Example 1;
Example 2;
80 N
600
s = 5 m
When a force that acts on an object moves the object through a distance in the direction of the force.
of a force and the distance traveled in the direction of the force.
WORK = FORCE X DISPLACEMENT W = F x s
W : work in Joule/J F : force in Newton/N s : displacement in meter/m
W = Fs
If, F = 40 N and s = 2 m Hence, W = 40 x 2 = 80 J
Force, F
s
W = Fs
= 80 cos 600 (5) = 80 (0.5) (5)
= 200 J
28 Example 3;
Example 4;
F = 600 N
S = 0.8 m
Energy
1. Energy is ...
2. Energy cannot be ...
3. Exist in various forms such as ………...………
………
4. Example of the energy transformation;
………
………
5. ………
Example :
………
T T
F = 30 N
h = 1.5 m
W = F s = F h
= 30 (1.5)
= 45.0 J
W = F s
= 600 x 0.8
= 480 J
It is the potential to do work.
created nor be destroyed.
potential energy, kinetic energy, electrical energy, sound energy, nuclear energy and chemical energy.
When we are running up a staircase the work done consists of energy change from Chemical Energy Kinetic Energy Potential Energy
The energy quantity consumed is equal to the work done.
If 100 J of work is done, it means 100 J of energy is consumed.
29 Work done and the change in kinetic energy
1. Kinetic energy is ………
2. Refer to the figure above,
3. Example 1; A small car of mass 100 kg is moving along a flat road. The resultant force on the car is 200 N.
a) What is its kinetic energy of the car after moving through 10 m?
b) What is its velocity after moving through 10 m?
Work done and gravitational potential energy
h = 1.5 m
1. Gravitational potential energy is………...
………
2. Refer to the figure above;
3. Example; If m = 10 kg s Force, F
Through, v2 = u2 +2as u = 0
and, as = ½ v2
energy of an object due to its position.
(possessed by an object due to its position in a gravitational field) W = Fs = mg h where, F = mg
So, Gravitational energy, Ep = mgh
energy of an object due to its motion.
Work = Fs = mas
= m ( ½ v2)
The formulae of Kinetic energy, Ek = ½ mv2
Solution : Given : m = 100 kg , F = 200 N a. Kinetic energy, Ek = Fs
= 200 x 10= 2000 J
b. Velocity, v ½ mv2 = 2000
v = 6.32 m s-1
W = 10 (10) 1.5
= 1500 J
Therefore Work done = 1500J And, Ep = 1500 J
30 Principle of conservation of energy
Carry out hands-on activity 2.10 on page 38 of the practical book.
To show the principle of conservation of energy.
1. Energy cannot be ………
………
2. Example : a thrown ball upwards will achieve a maximum height before changing its direction and falls
3. Example in calculation : A coconut falls from a tree from a height of 20 m. What is the velocity of coconut just before hitting the earth?
Power
1. Power is ………
2. A weightlifter lifts 180 kg of weights from the floor to a height of 2 m above his head in a time of 0.8 s. What is the power generated by the weightlifter during this time?
g = 9.8 ms-2)
31 Efficiency
1. Defined……..……….
2. Formulae of efficiency :
3. Analogy of efficiency;
Energy transformation
4. Example; An electric motor in a toy crane can lift a 0.12 kg weight through a height of 0.4 m in 5 s. During this time, the batteries supply 0.8 J of energy to the motor. Calculate (a) The useful of output of the motor.
(b) The efficiency of the motor
Carry out hands-on activity 2.11 on page 39 of the practical book to measure the power.
Device/
mechine Device/
mechine
as the percentage of the energy input that is transformed into useful energy.
%
100
Energyinput output energy Useful
Efficiency
unwanted energy
Energy input, Einput Useful energy, Eoutput
Solution : Given : m = 0.12 kg, s= 0.4 m, t = 5 s, Einput = 0.8 J (a) Eoutput = ?
Eoutput = F x s
= (0.12 x 10) x 0.4 = 0.48 J
(b) Efficiency = ?
Efficiency x100% E
E
input output
x 100%
0.80
0.48 60%
32 Exercise 2.10
1. What is the work done by a man when he pushes a box with a force of 90 N through a distance of 10 m? State the amount of energy transferred from the man to the force.
2. A sales assistant at a shop transfers 50 tins of milk powder from the floor to the top shelf.
Each tin has a mass of 3.0 kg and the height of thee top shelf is 1.5 m.
(a) Calculate the total work done by the sales assistant.
(b) What is his power if he completes this work in 250 s?
2.11 APPRECIATING THE IMPORTANCE OF MAXIMISING THE EFFICIENCY
OF DEVICES
1. During the process of transformation the input energy to the useful output
energy,………..
2. .………..
3. ………
Example of wasting the energy;
………..………
Input enegy output
from the petrol energy
……… ………. ……… ……….
..……….. ……….. ……….. ……….
..……….. ………. ………. ……….
some of energy transformed into unwanted forms of energy.
The efficiency of energy converters is always less than 100%.
The unwanted energy produced in the device goes to waste.
Kinetic energy
Energy loss due to Energy loss Energy loss Energy loss due to friction at friction in as heat as sound other parts in the
moving parts engine
W = F s The energy transferred to the force = 900 J = 90 x 10
= 900 J
m = 3.0 x 50 = 150 kg h = 1.5 m W = mhg = 150 x 9.8 x 1.5
= 2205 J
P = t W
= 250
2205 = 8.82 w
33
4. The world we are living in face acute shortage of energy.
5. It is very important that a device makes
………
Ways of increasing the efficiency of devices
1. Heat engines ………..………
1. The electrical devices increase the efficiency……….……
2. Proper management …...………
- choose the capacity according to the size of the family.
- installed away from source of heat and direct sunlight.
- the door must always be shut tight.
- more economical use a large capacity refrigerator.
- use manual defrost consumption.
Washing machines
- use a front loading as such more economical on water and electricity - front loading use less detergent as compared to a top loading machine.
the best possible use of the input energy.
Engine must be designed with the capability to produce greater amount of mechanical work.
Light Fittings
- replace filament light bulb with fluorescent lamps which have higher efficiency.
- use a lamp with a reflector so that the illumination can be directed to specific areas of the user.
Air-conditioners.
- choose a model with a high efficiency.
- accommodate the power of air-conditioner and the size of the room
- Ensure that the room totally close so that the temperature in the room can be maintained.
34 2.12 UNDERSTANDING ELASTICITY
Carry out Hands-on activity 2.12 page 40 of the practical book.
1. Elasticity is ………...
………
2. Forces between atoms ………..
………
3. Forces between atoms in equilibrium condition
Explanation :
………
………
………
4. Forces between atoms in compression
Explanation ;
………
………
………
5. Forces between atoms in tension
force of attraction
stretching force stretching force Explanation ;
………
………
………
Force of repulsion
Force of attraction
Force of repulsion
compressive force compressive force
Force of repulsion Force of repulsion
1. Force of attraction takes effect.
2. When the compressive force is removed, force of repulsion between the atoms pushes the atom back to their equilibrium positions.
1. Force of repulsion takes effect.
2. When the compressive force is removed, force of repulsion between the atoms pushes
the atom back to their equilibrium positions.
the property of an object that enables it to return its original shape and dimensions after an applied external force is removed.
The property of elasticity is caused by the existence of forces of repulsion and attraction between molecules in the solid material.
1. The atoms are separated by a distance called the equilibrium distance and vibrate at it position.
2. Force of repulsion = Force of attraction
35
Carry out Experiment 2.4 on page 41 of the practical book
To investigate the relationship between force and extension of a spring Hooke’s Law
1. Hooke’s Law states ………
………
2. Elastic limit of a spring is defined……….
………
3. The spring is said to have a permanent extension,...………
………
………
4. The elastic limit is not exceeded,……….…………
………
When the spring obey Hooke’s Law.
The mathematical expression for Hooke’s Law is : F x force provided that the elastic limit is not exceeded.
as the maximum force that can be applied to spring such that the spring will return to its original length when the force released.
when the length of the
spring longer than the original length even though the force acts was released and the elastic limit is exceeded.
36
Example 1; A spring has an original length of 15 cm. With a load of mass 200 g attached, the length of the spring is extend to 20 cm.
a. Calculate the spring constant.
b. What is the length of the spring when the load is in increased
Other situation where the spring extended and compressed
Given : lo = 15 cm, m = 200 g , F = 2.0 N, l = 20 cm x = 5 cm a. k = ?, k = 0.4Ncm 1
2.05
Fx
The graph shows the relationship between the stretching force, F and the spring extension, x.
(a) Calculate the spring constant of P and Q.
(b) Using the graph, determine the stretching force acts to spring P and
spring Q, when their extension are 0.5 cm Solution
a. Spring constant, k = gradient of graph kP = 15.79 N cm 1 ( extrapolation of graph P)
FQ = 3.0 N
37
Relationship between work and elastic potential energy Graph F against x
Example ;
Factors that effect elasticity
Hands-on activity 2.13 on page 42 the practical book to investigate the factors that affect elasticity.
Type of material different same same same Diameter of spring wire same different same same Diameter of spring same same different same
Length of spring same Same same different Summarise the four factors that affect elasticity
Factor Change in factor Effect on elasticity
Length Shorter spring Less elastic
Longer spring More elastic
Diameter of spring
Smaller diameter Less elastic
Larger diameter More elastic
Diameter of spring wire
Smaller diameter More elastic
Larger diameter Less elastic
Type of material the elasticity changes with the type of materials
x / cm F/N
F
x
Area under the graph = work done = ½ Fx So, Elastic potential energy = ½ Fx
15 cm
5 kg
8 cm
x = 15 – 8 = 7 cm = 0.07 m
Force act to the spring, F = 5 x 10 = 50 N
Elastic potential energy = ½ Fx
= ½ 50 (0.07) = 1.75 J
38 Exercise 2.12
1. A 6 N force on a spring produces an extension of 2 cm. What is the extension when the force is increased to 18 N? State any assumption you made in calculating your answer.
2. If a 20 N force extends a spring from 5 cm to 9 cm, (a) what is the force constant of the spring?
(b) Calculate the elastic potential energy stored in the spring.
Reinforcement Chapter 2
2. In an inelastic collision, which of the following quantities remains
3. Calculate the weight of a stone with mass 60 g on the surface of the
To solve the problem, determine the spring constant to use the formula F = k x F = 6 N , x = 2 cm
39
5. Which of the following diagrams shows a body moving at constant velocity?
A. 2 N 2N
B. 12 N 7 N C. 12 N 14 N D. 20 N 17 N
6. The graph below shows the motion of a trolley with mass 1.5 kg.
7. This figure shows an aircraft flying in the air.
8. m = 0.3 kg 5 m
What is the momentum of the stone just before it hits the ground?
A. 0.15 kg m s-1
Calculate the time needed for the iron ball to land.