Physic Chap 2 Note

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    Linear Motion

    Linear motion is the motion in 1 dimension (1-D) or the motion in a straight line.

    Distance

    1) The distance traveled by an object is the total length that is traveled by thatobject.

    2) Distance is a scalar quantity.

    3) The SI unit of distance is m (metre).

    Displacement

    1) Displacement of an object from a point of reference, O is the shortest distanceof the object from point O in a specific direction.

    2) Displacement is a vector quantity.

    3) The SI unit of displacement is m (metre).

    Distance vs Displacement

    Distance travelled = 200m

    Displacement = 120 m, in the direction of Northeast

    Speed

    1. Speed is the rate of change in distance. It is a measure of how fast the distance

    change in a movement.

    2. Speed is a scalar quantity.3. The SI unit of speed is m/s (metre per second)

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    Equation of Speed

    Velocity

    1. Velocity is define as the rate of displacement change. It is the measure of how

    fast the displacement change of a moving object.2. Velocity is a vector quantity.

    3. The unit of displacemnet is m/s (metre per second)

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    Equation of velocity

    Positive or Negative Sign of Velocity

    1. In velocity, the positive/negative sign indicates direction.

    2. You can take any direction as positive and the opposite as negative.

    3. For a linear motion, normally we take the motion to the right as positive andhence the motion to the left as negative.

    Acceleration

    Acceleration is the rate of velocity change.Acceleration is a vector quantity. It is ameasure of how fast the velocity change.

    Acceleration is a vector quantity.

    The unit of acceleration is ms-2.

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    Equation

    Additional Notes

    An object moves with a constant velocity if the magnitude and direction of the motion is

    always constant.An object experiences changes in velocity if

    the magnitude of velocity changes

    the direction of the motion changes.

    An object that experiences changes in velocity is said to have acceleration.

    An object traveling with a constant acceleration, a, if the velocity changes at a constantrate.

    Equation of Uniform Acceleration

    Most of the motion problems can be solved by the following equations. Therefore, make

    sure that you memorise all of them.

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    How we know when to use the equation?

    There are 3 types of motion:

    motion with uniform velocity

    motion with uniform acceleration

    motion with changing acceleration

    The 4 equations are used when the motion is uniform acceleration.

    Motion with changing acceleration is not in SPM Physics syllabus. It will be discussed in

    Form 5 add maths.

    Ticker Timer

    A ticker-timer consists of an electrical vibrator which vibrates 50 times per second. This

    enables it to make 50 dots per second on a ticker-tape being pulled through it. The timeinterval between two adjacent dots on the ticker-tape is called one tick. One tick is equal

    to 1/50 s or 0.02 s.

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    Uniform Velocity

    Thedistance of the dots is equally distributed.

    All lengths of tape in the chart are of equal length.

    The object is moving at a uniform velocity.

    Uniform Acceleration

    The

    distance between the dots increases uniformly.

    The length of the strips of tape in the chart increase uniformly.

    The velocity of the object is increasing uniformly, i.e. the object is moving at a

    constant acceleration.

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    Uniform Deceleration

    The

    distance between the dots decreases uniformly.

    The length of the strips of tape in the chart decreases uniformly.

    The velocity of the object is decreasing uniformly, i.e. the object is decelerating

    uniformly.

    Finding Velocity

    Velocity of a motion can be determined by using ticker tape through the following

    equation:

    Caution!:

    t is time taken from the first dot to the last dot of the distance measured.

    Example 1

    Diagram 2.4 shows a strip of ticker tape that was pulled through a ticker tape timer that

    vibrated at 50 times a second. What is the

    a. time taken from the first dot to the last dot?

    b. average velocity of the object that is represented by the ticker tape?

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    Answer:

    a. There are 15 ticks from the first dot to the last dot, hence

    Time taken = 15 0.02s = 0.3sb. Distance travelled = 15cm

    Example 1

    The ticker-tape in figure above was produced by a toy car moving down a tilted runway.

    If the ticker-tape timer produced 50 dots per second, find the acceleration of the toy car.

    Answer:

    In order to find the acceleration, we need to determine the initial velocity, the final

    velocity and the time taken for the velocity change.Initial velocity,

    u = \frac{s}{t} = \frac{{3cm}}

    {{0.02s}} = 150cms^{ - 1}

    Time taken for the velocity change,t = (0.5 + 4 + 0.5) ticks = 5 ticks

    t = 5 0.02s = 0.1s

    Example 2

    A trolley is pushed up a slope. Diagram above shows ticker tape chart that show the

    movement of the trolley. Every section of the tape contains 5 ticks. If the ticker-tape

    timer produced 50 dots per second, determine the acceleration of the trolley.

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    Answer:

    In order to find the acceleration, we need to determine the initial velocity, the final

    velocity and the time taken for the velocity change.Initial velocity,

    Time taken for the velocity change,

    t = (2.5 + 5 + 5 + 5 + 2.5) ticks = 40 ticks

    t = 40 0.02s = 0.8s

    Displacement - Time Graph

    In a Displacement-Time Graph, the gradient of the graph is equal to the velocity ofmotion.

    Analysing Displacement - Time Graph

    Gradient = 0Hence, velocity = 0

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    Gradient is constant,

    hence, velocity is Uniform

    Gradient is negative and constant, hence

    velocity is uniform and in opposite direction

    Gradient is increasing, hence velocity is

    increasing.

    Gradient is decreasing, hence velocity is

    decreasing.

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    Velocity - Time Graph

    The gradient of the velocity-time gradient gives a value of the changing rate invelocity, which is the acceleration of the object.

    The area below the velocity-time graph gives a value of the object's displacement.

    Analysing Velocity-Time Graph

    Uniform velocity

    Uniform acceleration

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    Increasing acceleration

    Uniform deceleration

    Decreasing acceleration

    Converting a Velocity-Time graph to Acceleration-Time graph

    In order to convert a velocity-time graph to acceleration time graph, we need to find the

    gradient of the velocity time graph and plot it in the acceleration-time graph.Free fallingis a motion under gravitational force as the only force acting on the moving object. In

    SPM, you need to know the graphs of free falling of the following movement

    1. Launching object upward.2. Dropping Object from High Place

    3. Object Falling and Bounce Back

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    Launching Object Upward

    Velocity-Time Graph Acceleration-Time Graph

    Dropping Object from High Place

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    Velocity-Time Graph Acceleration-Time Graph

    Object Falling and Bounce Back

    Velocity-Time Graph Acceleration-Time Graph

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    Mass

    Mass is defined as the amount of matter. The SI unit of mass is kilogram (kg)

    Mass is a scalar quantity.

    InertiaInertia is the property of a body that tends to maintain its state of motion.

    Newton's First Law

    In the absence of external forces, an object at rest remains at rest and an object in motion

    continues in motion with a constant velocity (that is, with a constant speed in a straightline).

    Jerking a Card

    When the cardboard is jerked quickly, the coin will fall into the glass.

    Explanation:

    The inertia of the coin resists the change of its initial state, which is stationary.

    As a result, the coin does not move with the cardboard and falls into the glassbecause of gravity.

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    Pulling a Book

    When the book is pulled out, the books on top will fall downwards.

    Explanation:

    Inertia tries to oppose the change to the stationary situation, that is, when the book

    is pulled out, the books on top do not follow suit.

    Pulling a Thread

    1. Pull slowly - Thread A will snap.

    Explanation:

    Tension of thread A is higher than string B.

    Tension at A = Weight of the load + Pulling Force

    2. Yank quickly - Thread B will snap.

    Explanation:

    The inertia of the load prevents the force from being transmitted to thread A,hence causing thread B to snap.

    Larger Mass - Greater Inertia

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    Bucket filled with sand is more difficult to be moved. It's also more difficult to be

    stopped from swinging.

    Explanation:

    Object with more mass offers a greater resistance to change from its state of

    motion.

    Object with larger mass has larger inertia to resist the attempt to change the stateof motion.

    Empty cart is easier to be moved

    An empty cart is easier to be moved compare with a cart full with load. This is because acart with larger mass has larger inertia to resist the attempt to change the state of motion.

    Momentum

    Momentum is defined as the product of mass and velocity.Momentum is a vector quantity. It has both magnitude and direction.

    The SI unit of momentum is kgms-1

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    Formula:

    Example 1

    A student releases a ball with mass of 2 kg from a height of 5 m from the ground. What

    would be the momentum of the ball just before it hits the ground?

    Answer:

    In order to find the momentum, we need to know the mass and the velocity of the ball

    right before it hits the ground.

    It's given that the mass, m = 2kg.The velocity is not given directly. However, we can determine the velocity, v, by using

    the linear equation of uniform acceleration.

    This is a free falling motion,

    The initial velocity, u = 0The acceleration, a = gravirational acceleration, g = 10ms-2

    The dispacement, s = high = 50m.

    The final velocity = ?From the equation

    v2 = u2 + 2as

    v2 = (0)2 + 2(10)(5)

    v = 10ms

    -1

    The momentum,

    p = mv =(2)(10) = 20 kgms-1

    Principle of Conservation of Momentum

    The principle of conservation of momentum states that in a system make out of objects

    that react (collide or explode), the total momentum is constant if no external force is

    acted upon the system.Sum of Momentum Before Reaction

    = Sum of Momentum After Reaction

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    Formula

    Example 1 - Both Object are in the Same Direction before Collision

    A Car A of mass 600 kg moving at 40 ms-1 collides with a car B of mass 800 kg moving

    at 20 ms-1 in the same direction. If car B moves forwards at 30 ms-1 by the impact, what is

    the velocity, v, of the car A immediately after the crash?

    Answer:m1 = 600kg

    m2 = 800kgu1 = 40 ms

    -1

    u2 = 20 ms-1

    v1 = ?v2 = 30 ms

    -1

    According to the principle of conservation of momentum,

    m1u1 + m2u2 = m1v1 + m2v2(600)(40) + (800)(20) = (600)v1 + (800)(30)40000 = 600v1 + 24000

    600v1 = 16000v1 = 26.67 ms-1

    Example 2 - Both Object are in opposite direction Before Collision

    A 0.50kg ball traveling at 6.0 ms-1 collides head-on with a 1.0 kg ball moving in theopposite direction at a speed of 12.0 ms-1. The 0.50kg ball moves backward at 14.0 ms-1

    after the collision. Find the velocity of the second ball after collision.

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    Answer:

    m1 = 0.5 kg

    m2 = 1.0 kgu1 = 6.0 ms

    -1

    u2 = -12.0 ms-1

    v1 = -14.0 ms-1

    v2 = ?

    (IMPORTANT: velocity is negative when the object move in opposite direction)

    According to the principle of conservation of momentum,m1u1 + m2u2 = m1v1 + m2v2(0.5)(6) + (1.0)(-12) = (0.5)(-14) + (1.0)v2-9 = - 7 + 1v2v2 = -2 ms

    -1

    Explosion

    Before explosion both object stick togetherand at rest.

    After collision, both object move at oppositedirection.

    Total Momentum before collision Is zero Total Momentum after collision :

    m1v1 + m2v2

    From the law of conservation of momentum:

    Total Momentum Before collision = Total Momentum after collision

    0 = m1v1 + m2v2

    m1v1 = - m2v2

    (-ve sign means opposite direction)

    Examples or Application of Conservation of Momentum in Explosion

    1. Fire a pistol or rifle2. Launching a rocket

    3. Application in jet engine

    4. Fan boat

    Example 3

    A man fires a rifle which has mass of 2.5 kg. If the mass of the bullet is 10 g and it

    reaches a velocity of 250 m/s after shooting, what is the recoil velocity of the pistol?

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    Answer

    This is a typical question of explosion.

    m1 = 2.5 kgm2 = 0.01 kg

    u1 = 0 ms-1

    u2 = 0 ms-1

    v1 = ?

    v2 = 250 ms-1

    By using the equation of conservation of momentum principle0 = m1v1 + m2v20 = (2.5)v1 + (0.01)(250)

    (2.5)v1 = -2.5v1 = -1 ms-1

    Elastic Collision

    Elastic collision is the collision where the kinetic energy is conserved after the collision.

    Total Kinetic Energy before Collision= Total Kinetic Energy after Collision

    Additional notes:

    In an elastic collision, the 2 objects separated right after the collision, and

    the momentum is conserved after the collision.

    Total energy is conserved after the collision.

    Inelastic Collision

    Inelastic collision is the collision where the kinetic energy is not conserved after the

    collision.Additional notes:

    In a perfectly elastic collision, the 2 objects attach together after the collision,

    and

    the momentum is also conserved after the collision.

    Total energy is conserved after the collision.

    Example 1 - Perfectly Inelastic Collision

    A lorry of mass 8000kg is moving with a velocity of 30 ms-1. The lorry is thenaccidentally collides with a car of mass 1500kg moving in the same direction with a

    velocity of 20 ms-1. After the collision, both the vehicles attach together and move with a

    speed of velocity v. Find the value of v.

    Answer:

    (IMPORTANT: When 2 object attach together, they move with same speed.)m1 = 8000kg

    m2 = 1500kg

    u1 = 30 ms-1

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    u2 = 20 ms-1

    v1 = v

    v2 = vAccording to the principle of conservation of momentum,

    m1u1 + m2u2 = m1v1 + m2v2

    (8,000)(30) + (1,500)(20) = (8,000)v+ (1,500)v270,000 = 9500v

    v = 28.42 ms-1

    Rocket

    1. 1Mixture of hydrogen and oxygen fuels burn in the combustion chamber.2. Hot gases are expelled through the exhausts at very high speed .

    3. The high-speed hot gas produce a high momentum backwards.

    4. By conservation of momentum, an equal and opposite momentum is produced and

    acted on the rocket, pushing the rocket upwards.

    Jet Engine

    1. Air is taken in from the front and is compressed by the compressor.

    2. Fuel is injected and burnt with the compressed air in the combustion chamber.

    3. The hot gas is forced through the engine to turn the turbine blade, which turns thecompressor.

    4. High-speed hot gases are ejected from the back with high momentum.

    5. This produces an equal and opposite momentum to push the jet plane forward.

    Force

    1. A force is push or pull exerted on an object.2. Force is a vector quantity that has magnitude and direction.

    3. The unit of force is Newton ( or kgms-2).

    Unbalanced Force/ Resultant Force

    When the forces acting on an object are not balanced, there must be a net force acting on

    it. The net force is known as the unbalanced force or the resultant force.

    When a force acts on an object, the effect can change the

    1. size,

    2. shape,3. stationary state,

    4. speed and

    5. direction of the object.

    Newton's Second Law

    The rate of change of momentum of a body is directly proportional to the resultant force

    acting on the body and is in the same direction.

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    Implication:

    When there is resultant force acting on an object, the object will accelerate (moving

    faster, moving slower or change direction).

    Formula of Force

    From Newton's Second Law, we can derived the equation

    (IMPORTANT: F Must be the net force)

    Summary of Newton's 1st Law and 2nd Law

    Newton's First Law:

    When there is no net force acting on an object, the object is either stationary or move

    with constant speed in a straight line.

    Newton's Second Law:

    When there is a net force acting on an object, the object will accelerate.

    Example 1

    A box of mass 150kg is placed on a horizontal floor with a smooth surface; find theacceleration of the box when a 300N force is acting on the box horizontally.

    Answer:

    F = ma

    (300) = (150)aa = 2 ms-2

    Example 2

    A object of mass 50kg is placed on a horizontal floor with a smooth surface. If thevelocity of the object changes from stationary to 25.0 m/s in 5 seconds when is acted by a

    force, find the magnitude of the force that is acting?Answer:We know that we can find the magnitude of a force by using the formula F = ma. The

    mass m is already given in the question, but the acceleration is not give directly.

    We can determine the acceleration from the formula .From the formulaF = ma = (50)(5) = 250N

    The force acting on the box is 250N.

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    Impulse

    Impulse is defined as the product of the force (F) acting on an object and the time of

    action (t). Impulse exerted on an object is equal to the momentum change of the object.Impulse is a vector quantity.

    Formula of impulseImpulse is the product of force and time.

    Impulse = F t

    Impulse = momentum changeImpulse = mv mu

    Impulsive Force

    Impulsive force is defined as the rate of change of momentum in a reaction.

    It is a force which acts on an object for a very short interval during a collision orexplosion.

    Effects of impulse vs Force

    A force determines the acceleration (rate of velocity change) of an object. A

    greater force produces a higher acceleration. An impulse determines the velocity change of an object. A greater impulse yield a

    higher velocity change.

    Examples Involving Impulsive Force

    Playing football

    Playing badminton Playing tennis

    Playing golf

    Playing baseball

    Long Jump

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    1. The long jump pit is filled with sand to increase the reaction time when atlete land

    on it.

    2. This is to reduce the impulsive force acts on the leg of the atlete becauseimpulsive force is inversely proportional to the reaction time.

    High Jump

    (This image is licenced under the GNU Free Document Licence. The original file is from

    the Wikipedia.org.)

    During a high jump, a high jumper will land on a thick, soft mattress after the

    jump.

    This is to increase the reaction time and hence reduces the impulsive force acting

    on the high jumper.

    JumpingA jumper bends his/her leg during landing. This is to increase the reaction time and hencereduce the impact of impulsive force acting on the leg of the jumper.

    Crumble Zone

    The crumple zone increases the reaction time of collision during an accident.

    This causes the impulsive force to be reduced and hence reduces the risk of injuries.

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    Seat Belt

    Prevent the driver and passengers from being flung forward or thrown out of the carduring an emergency break.

    Airbag

    The inflated airbag during an accident acts as a cushion to lessen the impact when the

    driver flings forward hitting the steering wheel or dashboard.

    Head Rest

    Reduce neck injury when driver and passengers are thrown backwards when the car is

    banged from backward.

    Windscreen

    Shatter-proof glass is used so that it will not break into small pieces when broken. This

    may reduce injuries caused by scattered glass.

    Padded Dashboard

    Cover with soft material. This may increases the reaction time and hence reduce the

    impulsive force when passenger knocking on it in accident.

    Collapsible Steering Columns

    The steering will swing away from drivers chest during collision. This may reduce the

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    impulsive force acting on the driver.

    Anti-lock Braking System (ABS)

    Prevent the wheels from locking when brake applied suddenly by adjusting the pressure

    of the brake fluid. This can prevents the car from skidding.

    Bumper

    Made of elastic material so that it can increases the reaction time and hence reduces the

    impulsive force caused by collision.

    Passenger Safety Cell

    The body of the car is made from strong, rigid stell cage.

    This may prevent the car from collapsing on the passengers during a car crash.

    Gravitational Field

    A gravitational field as a region in which an object experiences a force due togravitational attraction.

    Gravitational Field Strength

    The gravitational field strength at a point in the gravitational field is the gravitationalforce acting on a mass of 1 kg placed at that point.

    The unit of gravitational field strength is N/kg.

    The gravitational field strength is denoted by the symbol "g".

    Gravitational Field Strength Formula

    Gravitational Acceleration

    The gravitational acceleration is the acceleration of an object due to the pull of the

    gravitational force.

    The unit of gravitational acceleration is ms-2

    Gravitational acceleratio is also denoted by the symbol "g".

    Symbol: g

    Important notes:

    Gravitational acceleration does not depend on the mass of the moving object.

    The magnitude of gravitational acceleration is taken to be 10ms-2.

    Gravitational Field Strength vs. Gravitational Acceleration

    Both the gravitational field strength and gravitational acceleration have the

    symbol, g and the same value (10ms-2) on the surface of the earth.

    When considering a body falling freely, the g is the gravitational acceleration.

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    When considering objects at rest, g is the Earths gravitational field strength

    acting on it.

    Weight

    The weight of an object is defined as the gravitational force acting on the object.The SI unit of weight is Newton (N)

    Differences between Weight and Mass

    Weight Mass

    Depends on the gravitational fieldstrength

    Independent from the gravitational fieldstrength

    Vector quantity Scalar Quantity

    Unit Newton (N) Unit: Kilogram (kg)

    Free Falling

    1. Free falling is a motion under force of gravity as the only force acting on themoving object.

    2. Practically, free falling can only take place in vacuum.

    Gravitational Acceleration

    1. The gravitational acceleration is the acceleration of an object due to the pull of thegravitational force. It has the unit of ms-2

    2. The symbol of gravitational acceleration is " g ".

    3. Gravitational acceleration does not depend on the mass of the moving object.

    4. The magnitude of gravitational acceleration is taken to be 10ms-2.

    Gravitational Field Strength vs. Gravitational Acceleration

    1. Both the gravitational field strength and gravitational acceleration have the

    symbol, g and the same value (10ms

    -2

    ) on the surface of the earth.2. When considering a body falling freely, the g is the gravitational acceleration.3. When considering objects at rest, g is the Earths gravitational field strength

    acting on it.

    Vector and Scalar Quantity

    A scalar quantity is a quantity which can be fully described by magnitude only.

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    A vector quantity is a quantity which is fully described by both magnitude and direction.

    Vector Diagram

    The arrow shows the direction of the vector.

    The length representing the magnitude of the vector.

    Equal Vector

    Two vectors A and B may be defined to be equal if they have the same magnitude and

    point in the same direction.

    Case of Free Falling 1 - Falling from High Place

    When an object is released from a high place,

    1. its initial velocity, u = 0.

    2. its acceleration is equal to the gravitational acceleration, g, which taken to be10ms-2 in SPM.

    3. the displacement is the of the object when it reaches the ground is equal to theinitial height of the object, h.

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    Case of Free Falling 2 - Launching Object Upward

    If an object is launched up vertically,

    1. the acceleration = -g (-10ms-2)

    2. the velocity become zero when the object reaches the highest point.

    3. the displacement of the object at highest point is equal to the vertical height ofobject, h

    4. the time taken for the object to move to the maximum height = the time taken for

    the object to fall from the maximum point to its initial position.

    Vector Addition - Triangle Method

    Join the tail of the 2nd vector to the head of the 1st vector. Normally the resultant vector

    is marked with double arrow.

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    Vector Addition - Parallelogram Method

    Join the tail of the 2nd vector to the tail of the 1st vector. Normally the resultant vector ismarked with double arrow.

    Addition of 2 Perpendicular Vectors

    If 2 vectors (a and b) are perpendicular to each others, the magnitude and direction of the

    resultant vector can be determined by the following equation.

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    Example 2

    Diagram above shows that four forces of magnitude 2N, 4N, 5N and 8N are acting onpoint O. All the forces are perpendicular to each others. What is the magnitude of the

    resulatant force that acts on point O?

    Answer:

    The resultant force of the horizntal component = 5 - 2 = 3N to the right

    The resultant force of the vertical component = 8 - 4 = 4N acting downward.Therefore, the magtitude of these 2 force components,

    Vector Resolution

    A vector can be resolve into 2 component which is perpendicular to each others.

    Example 1

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    Diagram above shows a lorry pulling a log with an iron cable. If the tension of the cable

    is 3000N and the friction between the log and the ground is 500N, find the horizontal

    force that acting on the log.

    Answer:Horizontal component of the tension = 3000 cos30o =2598N

    Friction = 500N

    Resultant horizontal force = 2598N - 500N =2098N

    Example 2

    Diagram above shows two forces of magnitude 25N are acting on an object of mass 2kg.Find the acceleration of object P, in ms-2.

    Answer:

    Horizontal component of the forces = 25cos45o + 25cos45o = 35.36N

    Vertical component of the forces = 25sin45o - 25sin45o = 0NThe acceleration of the object can be determined by the equation

    F = ma

    (35.36) = (2)aa = 17.68 ms-2

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    Inclined Plane

    Weight component along the plane = Wsin.

    Weight component perpendicular to the plane = Wcos.

    Example 3

    A block of mass 2 kg is pulling along a plane by a 20N force as shown in diagram above.

    Given that the fiction between block and the plane is 2N, find the magnitude of the

    resultant force parallel to the plane.Answer:

    First of all, let's examine all the forces or component of forces acting along the plane.The force pulling the block, F = 20N

    The frictional force Ffric = 2N

    The weight component along the plane = 20sin30o = 10NThe resultant force along the plane = 20 - 2 - 10 = 8N

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    Vectors in Equilibrium

    When 3 vectors are in equilibrium, the resultant vector = 0. After joining all the vectors

    tail to head, the head of the last vector will join to the tail of the first vector.

    Forces in equilibrium

    Forces are in equilibrium means the resultant force in all directions are zero.

    When the forces acting on an object are balanced, they cancel each other out. The net

    force is zero.

    Effect :

    an object at rest is continuely at rest [ velocity = 0]

    a moving object will move at constant velocity [ a = 0]

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    Example 1

    Diagram above shows a load of mass 500g is hung on a string C, which is tied to 2 otherstrings A and B. Find the tension of string A.

    Answer:

    Tension of string C, TC = weight of the load = 5N

    All forces in the system are in equilibrium, hence

    Vertical component of tension A (TA) = TCTAcos60

    o = TCTA = TC/cos60

    o

    TA = 5/cos60o = 10N

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    Work

    1. Work done by a constant force is given by the product of the force and the

    distance moved in the direction of the force.2. The unit of Nm(Newton metre) or J(Joule).

    3. Work is a scalar quantity.

    Equation of Work

    When the direction of force and motion are same, = 0o, therefore cos = 1Work done,

    W = F s

    Example 1

    A force of 50 N acts on the block at the angle shown in the diagram. The block moves a

    horizontal distance of 3.0 m. Calculate the work being done by the force.

    Answer:

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    Work done,

    W = F s cos

    W = 50 3.0 cos30o = 129.9J

    Example 2

    Diagram above shows a 10N force is pulling a metal. The friction between the block and

    the floor is 5N. If the distance travelled by the metal block is 2m, find

    a. the work done by the pulling force

    b. the work done by the frictional force

    Asnwer:

    (a) The force is in the same direction of the motion. Work done by the pulling force,

    W = F s = (10)(2) = 20J

    (b) The force is not in the same direction of motion, work done by the frictional force

    W = F s cos180o= (5)(2)(-1) = -10J

    Work Done Against the Force of Gravity

    Example 3

    Ranjit runs up a staircase of 35 steps. Each steps is 15cm in height. Given that Ranjit's

    mass is 45kg, find the work done by Ranjit to reach the top of the staircase.

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    Answer:

    In this case, Ranjit does work to overcome the gravity.

    Ranjit's mass = 45kgVertical height of the motion, h = 35 0.15

    Gravitational field strength, g = 10 ms-2Work done, W = ?

    W = mgh = (45)(10)(35 0.15) = 2362.5J

    Finding Work from Force-Displacement Graph

    In a Force-Displacement graph, work

    done is equal to the area in between the graph and the horizontal axis.

    Example 4

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    The graph above shows the force acting on a trolley of 5 kg mass over a distance of 10 m.

    Find the work done by the force to move the trolley.

    Answer:In a Force-Displacement graph, work done is equal to the area below the graph.Therefore, work done

    Energy

    Energy is defined as the capacity to do work. Work is done when energy

    is converted from one form to another.

    Nm or Joule(J)

    Gravitational Potential Energy

    Gravitational potential energy is the energy stored in an object as the

    result of its vertical position (i.e., height).

    Formula:

    Example 1

    A ball of 1kg mass is droppped from a height of 4m. What is the maximum

    kinetic energy possessed by the ball before it reached the ground?

    Answer

    According to the principle of conservation of energy, the amount of

    potential energy losses is equal to the amount of kinetic energy gain.

    Maximum kinetic energy

    = Maximum potentila energy losses

    = mgh = (1)(10)(4) = 40J

    Elastic Potential Energy

    Elastic potential energy is the energy stored in elastic materials as

    the result of their stretching or compressing.

    Formula:

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    Example 2

    Diagram above shows a spring with a load of mass 0.5kg. The extention

    of the spring is 6cm, find the energy stored in the spring.

    Answer:The energy stored in the spring is the elestic potential energy.

    Kinetic Energy

    Kinetic energy is the energy of motion.

    Equation of Kinetic Energy

    Example 1

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    Relationship between Energy and Work Done

    During a conversing of energy,

    Amount of Work Done = Amount of Energy Converted

    ExampleA trolley of 5 kg mass moving against friction of 5 N. Its velocity at A is 4ms-1 and itstops at B after 4 seconds. What is the work done to overcome the friction?

    Answer:

    In this case, kinetic energy is converted into heat energy due to the friction. The workdone to overcome the friction is equal to the amount of kinetic energy converted into heat

    energy, hence

    Power

    Power is the rate at which work is done, which means how fast a work is done.

    Formula:

    Example 1An electric motor takes 20 s to lift a box of mass 20kg to a height of 1.5 m. Find the

    amount of work done by the machine and hence find the power of the electric motor.

    Answer:

    Work done,

    W = mgh = (20)(10)(1.5) = 300J

    Efficiency

    The efficiency of a device is defined as the percentage of the energy input that is

    transformed into useful energy.

    Example

    In the example above, the input power is 100J/s, the desire output power (useful energy)

    is only 75J/s, the remaining power is lost as undisire output. Therefore, the efficiency of

    this machine is

    75/100 x 100% = 75%

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    Air Conditioner

    1. Switch off the air conditioner when not in use.

    2. Buy the air conditioner with suitable capacity according to the room size.3. Close all the doors and windows of the room to avoid the cool air in the room

    from flowing out.

    Refrigerator

    1. Always remember to close the door of refrigerator.

    2. Open the refrigerator only when necessarily.3. Always keep the cooling coil clean.

    4. Defrost the refrigerator regularly.

    5. Choose the refrigerator with capacity suitable for the family size.6. Refrigerator of large capacity is more efficient compare with refirgerator of small

    capacity.

    Lamp or Light Bulb

    1. Use fluorecent bulb rather than incandescent bulb. Fluorescent bulbs are muchmore efficient than incandescent bulbs.

    2. Use a lamp with reflector so that more light is directed towards thr desirable

    place.

    Washing Machine

    1. Use front-loading washing machine rather than top-loading wahing machine

    because it uses less water and electricity.2. Use washing machine only when you have sufficient clothes to be washed. Try to

    avoid washing small amount of clothes.

    Elasticity

    Elasticity is the ability of a sub-stance to recover its original shape and size after

    distortion.

    Forces Between Atoms

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    The intermolecular forces consist of an attractive force and a repulsive force.

    At the equilibrium distance d, the attractive force equal to the repulsive force.

    If the 2 atoms are brought closer, the repulsive force will dominate, produces a netrepulsive force between the atoms.

    If the 2 atoms are brought furhter, the attractive force will dominate, produces anet attractive force between the atoms.

    Graph of Forces Between 2 atoms

    x0 = Equilibrium Distance

    When the particles are compressed, x < x0, the attractive force between the particlesincreases.If the distance x exceeds the elastic limit, the attractive force will decreases.

    Hooke's Law

    Hooke's Law states that if a spring is not stretched beyond its elastic limit, the force that

    acts on it is directly proportional to the extension of the spring.

    Elastic Limit

    The elastic limit of a spring is defined as the maximum force that can be applied to aspring such that the spring will be able to be restored to its original length when the force

    is removed.

    Equation derived from Hooke's Law

    From Hook's Law, we can derived that

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    Spring Constant

    Spring constant is defined as the ratio of the force applied on a spring to the extension of

    the spring.It is a measure of the stiffness of a spring or elastic object.

    Graph of Streching Force - Extension

    Gradient = Spring constant

    Area below the graph = Work done

    F-x graph and spring constant

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    The higher the gradient, the greater the spring constant and the harder (stiffer) spring.

    For example, the stiffness of spring A is greater than spring B.

    Spring

    Arrangement in series: Arrangement in parallel:

    Extension = x number of springStiffness decreases

    Spring constant = k/number of spring

    Extension = x number of springStiffness increases

    Spring constant = k number of spring

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    Factors Affecting the Stiffness of Spring

    Stiffer Less stiff

    Material type of spring

    (A steel spring is stiffer than a copper spring)

    Diameter of wire of spring

    (The greater the diameter of the wire, the stiffer the spring)

    Diameter of the spring

    (The smaller the diameter of spring, the stiffer the spring)

    Length of the string

    (Shorter spring is stiffer)

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