Fizik Sempoi Crewz

1- Safuan a.k.a. SAM
AA110043
2- Amirul Hakim a.k.a. ZACK


3- Shafiq a.k.a. BOY

AA110388
4- Hilman a.k.a. BOBBY
5-Zul Azmi a.k.a Zul



Friday 9 September 2011

K!nEt!c EneRgY


Kinetic Energy

Kinetic energy is the energy of motion. An object that has motion - whether it is vertical or horizontal motion - has kinetic energy. There are many forms of kinetic energy - vibrational (the energy due to vibrational motion), rotational (the energy due to rotational motion), and translational (the energy due to motion from one location to another). To keep matters simple, we will focus upon translational kinetic energy. The amount of translational kinetic energy (from here on, the phrase kinetic energy will refer to translational kinetic energy) that an object has depends upon two variables: the mass (m) of the object and the speed (v) of the object. The following equation is used to represent the kinetic energy (KE) of an object.
where m = mass of object
v = speed of object
This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed. That means that for a twofold increase in speed, the kinetic energy will increase by a factor of four. For a threefold increase in speed, the kinetic energy will increase by a factor of nine. And for a fourfold increase in speed, the kinetic energy will increase by a factor of sixteen. The kinetic energy is dependent upon the square of the speed. As it is often said, an equation is not merely a recipe for algebraic problem solving, but also a guide to thinking about the relationship between quantities.
Kinetic energy is a scalar quantity; it does not have a direction. Unlike velocityaccelerationforce, and momentum, the kinetic energy of an object is completely described by magnitude alone. Like work and potential energy, the standard metric unit of measurement for kinetic energy is the Joule. As might be implied by the above equation, 1 Joule is equivalent to 1 kg*(m/s)^2.

Thursday 8 September 2011

RoTaTiOn aNd CiRcULaR MoTiOn

Mathematics of Circular Motion

There are three mathematical quantities that will be of primary interest to us as we analyze the motion of objects in circles. These three quantities are speed, acceleration and force. The speed of an object moving in a circle is given by the following equation.
The acceleration of an object moving in a circle can be determined by either two of the following equations.
The equation on the right (above) is derived from the equation on the left by the substitution of the expression for speed.
The net force (Fnet) acting upon an object moving in circular motion is directed inwards. While there may by more than one force acting upon the object, the vector sum of all of them should add up to the net force. In general, the inward force is larger than the outward force (if any) such that the outward force cancels and the unbalanced force is in the direction of the center of the circle. The net force is related to the acceleration of the object (as is always the case) and is thus given by the following three equations:
The equations in the middle (above) and on the right (above) are derived from the equation on the left by the substitution of the expressions for acceleration.
This set of circular motion equations can be used in two ways:
These two ways are illustrated below.

Equations as a Guide to Thinking

An equation expresses a mathematical relationship between the quantities present in that equation. For instance, the equation for Newton's second law identifies how acceleration is related to the net force and the mass of an object.
The relationship expressed by the equation is that the acceleration of an object is directly proportional to the net force acting upon it. In other words, the bigger the net force value is, the bigger that the acceleration value will be. As net force increases, the acceleration increases. In fact, if the net force were increased by a factor of 2, the equation would predict that the acceleration would increase by a factor of 2. Similarly, if the net force were decreased by a factor of 2, the equation would predict that the acceleration would decrease by a factor of 2.
Newton's second law equation also reveals the relationship between acceleration and mass. According to the equation, the acceleration of an object is inversely proportional to mass of the object. In other words, the bigger the mass value is, the smaller that the acceleration value will be. As mass increases, the acceleration decreases. In fact, if the mass were increased by a factor of 2, the equation would predict that the acceleration would decrease by a factor of 2. Similarly, if the mass were decreased by a factor of 2, the equation would predict that the acceleration would increase by a factor of 2.
As mentioned previously, equations allow for predictions to be made about the affect of an alteration of one quantity on a second quantity. Since the Newton's second law equation shows three quantities, each raised to the first power, the predictive ability of the equation is rather straightforward. The predictive ability of an equation becomes more complicated when one of the quantities included in the equation is raised to a power. For instance, consider the following equation relating the net force (Fnet) to the speed (v) of an object moving in uniform circular motion.
This equation shows that the net force required for an object to move in a circle is directly proportional to the square of the speed of the object. For a constant mass and radius, the Fnet is proportional to the speed2.
The factor by which the net force is altered is the square of the factor by which the speed is altered. Subsequently, if the speed of the object is doubled, the net force required for that object's circular motion is quadrupled. And if the speed of the object is halved (decreased by a factor of 2), the net force required is decreased by a factor of 4.


Equations as a Recipe for Problem-Solving

The mathematical equations presented above for the motion of objects in circles can be used to solve circular motion problems in which an unknown quantity must be determined. The process of solving a circular motion problem is much like any other problem in physics class. The process involves a careful reading of the problem, the identification of the known and required information in variable form, the selection of the relevant equation(s), substitution of known values into the equation, and finally algebraic manipulation of the equation to determine the answer. Consider the application of this process to the following two circular motion problems.

Sample Problem #1

A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.
The solution of this problem begins with the identification of the known and requested information.
Known Information:
m = 900 kg v = 10.0 m/s
R = 25.0 m
Requested Information:
a = ???? Fnet = ????
To determine the acceleration of the car, use the equation a = v2 / R. The solution is as follows:
a = v2 / R a = (10.0 m/s)2 / (25.0 m)
a = (100 m2/s2) / (25.0 m)
a = 4 m/s2
To determine the net force acting upon the car, use the equation Fnet = m•a. The solution is as follows.
Fnet = m • a Fnet = (900 kg) • (4 m/s2)
Fnet = 3600 N

 

Wednesday 7 September 2011

Chapter 3 K!NeM@T!cS


Kinematic Equations and Problem-Solving

The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic equations are:
In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value.
In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps:
  1. Construct an informative diagram of the physical situation.
  2. Identify and list the given information in variable form.
  3. Identify and list the unknown information in variable form.
  4. Identify and list the equation that will be used to determine unknown information from known information.
  5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information.
  6. Check your answer to insure that it is reasonable and mathematically correct.
The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below.

Example A

Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.)
The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of thestrategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below.
Diagram:Given:Find:
vi = +30.0 m/s
vf = 0 m/s

a = - 8.00 m/s2
d = ??
The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vfvia, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables.
Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.
(0 m/s)2 = (30.0 m/s)2 + 2*(-8.00 m/s2)*d
0 m2/s2 = 900 m2/s2 + (-16.0 m/s2)*d
(16.0 m/s2)*d = 900 m2/s2 - 0 m2/s2
(16.0 m/s2)*d = 900 m2/s2
d = (900 m2/s2)/ (16.0 m/s2)
d = (900 m2/s2)/ (16.0 m/s2)
d = 56.3 m
The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.)
The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!

Example B

Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/sfor a time of 4.10 seconds. Determine the displacement of Ben's car during this time period.
Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below.
Diagram:Given:Find:
vi = 0 m/s
t = 4.10 s

a = 6.00 m/s2
d = ??
The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables.
Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.
d = (0 m/s)*(4.1 s) + 0.5*(6.00 m/s2)*(4.10 s)2
d = (0 m) + 0.5*(6.00 m/s2)*(16.81 s2)
d = 0 m + 50.43 m
d = 50.4 m
The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.)
The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!
The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions.

Tuesday 6 September 2011

Soalan Baek Punye

QUESTION BETWEEN VECTOR AND KINEMATIC


1-      Hakim and Safuan are pulling a boat. Both of them pull at different direction which makes an angle of 45o to the direction of motion. They pull with the force of 60 N. The mass of the boat is 75 kg and the friction is 40 N. Calculate

i)                    Total force towards left
ii)                  Acceleration of the boat
 

60 N
                                               
45o                                          40 N
45o
 
60 N

Monday 15 August 2011

Chapter 4

DYNAMIC OF A PARTICLE

Newton's Law :
- 1st Law = Inertia
- 2nd Law = Motion / Force
- 3rd Law =Forces in Equilinrium

Definition :
Inertia * Tendency of an object to mantain its state of rest or uniform motion in a straight line unless it is acted        
              upon by an external force.
Motion * When a net forces acts on an object, the acceleration of the object is direectly proportional to the
               net force and has a magnitude that is inversely proportional to it mass.
Forces in Equilibirium * States that for every action, there is a reaction acting in the opposite direction with the
                                     same magnitude.

Sunday 3 July 2011

Chapter 1. Unit and Measurements.


1-       Base Quantity – is a physic quantity that can be measured without need calculation

2-       Derived Quantity – is obtained from derivation of the base quantity or in other
word a combination of two or more base quantity

3-      Prefix Modifiers of the Metric System :-

PREFIX
SYMBOL
FACTOR POWER

Tera

T
1012
Giga

G
109
Mega

M
106
Kilo

k
103
Hecto

h
102
Deca

da
101
Deci

d
10-1
Centi

c
10-2
Mili

m
10-3
Micro

ยต
10-6
Nano

n
10-9
Pico

P
10-12







4-      Dimension of derived quantity :-
DERIVED QUANTITY

DIMENSION
Volume

[L]3
Energy

[M][L]2[T]-2
Power

[M][L]-2[T]-3
Acceleration

[L][T]-2
Pressure

[M][L]-1[T]-2
Momentum

[M][L][T]-1