Inertia : Is a property of an object that is its tendency ...



Inertia : Is a property of an object that is its tendency to resist change in its velocity.

The mass of an object is a measure of its inertia.

If an object has a large mass it requires a large force to set it in motion or to change its velocity.

If an object has a large mass it requires a large force to stop it if it is moving.

FORCE : Force is defined as that which changes or tends to change the state of rest or uniform motion in a straight line of a body.

OR

Force produces an acceleration which causes a change in the velocity of the object

OR

A body remains in a state of rest or uniform motion in a straight line unless it is acted on by an external force.

It can be shown that

Force = mass x acceleration

F = m . a

S.I. Units ( Basic ) kg . m.s-2

Which is called Newton N

The force of gravity on a body is called the weight of the body

Weight = m . g ( g = 9.81 m.s-2 )

Vector quantity which acts vertically down.

The S.I. unit of weight is Newton.

NOTE 1: Force is a vector quantity which is in the same direction as the acceleration since

(

F = m . a

The two vectors Force and acceleration are in the same direction

2. Also

Weight =mass x acceleration due to gravity

( (

W = m . g

The two vectors in the above equation are W and g and act in the same direction. Therefore W is in the same direction as g i.e. acting vertically downward.

However the acceleration value a is only equal to 9.81 ms-2 if and only if the object is free falling in the vertically downward direction.

PROBLEM SHEET.

Question 1. A constant force acts on a

5 kg object and reduces its velocity from

7 m.s-1 to 3m.s-1, along the +x direction , in a time of 3 seconds. Calculate the value of the force.

Question 2. A 6 kg object is to be given an acceleration of 0.7 m.s-2 along the +x direction calculate the value of the force acting on it.

Question 3. Find the weight of the following masses (a) 10 kg (b) 60 kg

Question 4. Calculate the mass of a body which has a weight of 100 N.

Question 5. Calculate the force on a mass of 10 kg if its velocity is to be increased from 30 km.hr-1 to 60 km.hr-1, along the +x direction , in a time of 10 seconds.

Question 6. Explain the difference between the mass and the weight of a substance giving the S.I. units of each.

Question 7. A mass of 5 kg starts from rest at the origin. Two Forces act on the mass F1 which is a force of 15 N in the +x direction and F2 a force of 8N at +45o.

i) Calculate the total force acting on the object.

ii) Calculate the total acceleration.

iii) Calculate the total displacement in a time of 10s

iv) Calculate the final velocity after 10s.

Question 8. Repeat question 7 if a third force F3 20 N along the +y direction also acts on the mass.

Momentum. The momentum of a body is defined as the product of its mass multiplied by its velocity.

Momentum = mass x velocity

Symbol used is p

p = m . v

S.I. Units kg m.s-1

VECTOR in the same direction as the velocity

Conservation of momentum states that when two or more bodies act on each other the total momentum of the system before the action is equal to the total momentum after the action provided that no external forces act on the system.

Action is usually a collision between the bodies.

NOTE: Any question which relates the a collision between two or more bodies is solved using the conservation of momentum.

Example 1. Calculate the momentum of an object of mass of 500 g moving at a velocity of 20 m.s-1 due East.

Example 2. An object of mass 1 kg moving at velocity 20m.s-1 due east collides with a stationary object of mass

3 kg. If the two masses couple together calculate the common velocity after the collision.

Example 3. An object moving with a steady velocity of 4 m.s-1 along the + x axis collides with a second object of double the mass and moving at half the velocity of the first object. If the two masses couple together calculate the common velocity after the collision.

Example 4. An object of mass 4 kg moving in the + x direction with a velocity of 10 m.s-1 collides with an object of mass 1 kg moving in the – x direction with a velocity of 4 m.s-1. Assuming that the two masses couple together calculate the common velocity after the collision.

Example 5. An object of mass 4 kg traveling along the + x direction with a velocity of 20m.s-1 collides with an object of mass 2 kg traveling along the – x direction with a velocity of 15 m.s-1. The two objects bounce off one another and after the collision the larger mass has a velocity of

15 m.s-1 in the -x direction, calculate the velocity of the smaller mass after the collision.

NEWTONS LAWS OF MOTION

First Law.

Every body remains in a state of rest or uniform motion in a straight line unless acted upon by a force.

Second Law.

The rate of change of momentum is proportional to the applied force and takes place in the direction of the force.

Third Law.

When one body exerts a force on a second body, then the second body exerts an equal and opposite force on the first body.

Action and Reaction are equal and opposite.

Newton’s First Law and circular motion.

Newtons first law which is the definition of force states that a body will remain at rest or uniform motion if a straight line unless it is acted upon by an external force.

Circular motion or motion in a circle.

When a body moves in a circle then, there must according to Newton’s first law be some force acting on it which tends to deflect it from its tendency to move in a straight line.

[pic]

This force is directed towards the centre of the circle around which the object is moving and is called the CENTRIPETAL FORCE.

If a particle is moving around a circle of radius r (m) with a uniform speed

v (m.s-1), the time take for the object to complete one revolution of the circle is called the PERIODIC TIME or PERIOD. To distinguish from the general symbol for time t we use the symbol capital T.

[pic]

In one periodic time the distance travelled would be equal to the circumference of the circle therefore

[pic]

Note although the speed is constant the velocity is changing because the direction is changing, because the direction of the motion is changing.

Around a circle the direction of the velocity at any point along the circle is the direction of the tangent to the circle at that point.

Since the velocity is changing, due to the changing direction there is an acceleration.

Acceleration is a vector quantity with both a magnitude and a direction.

The magnitude of the centripetal acceleration can be shown to be

[pic]

The direction of the acceleration can be shown be towards the centre of the circle around which the object is moving.

Given [pic]

then [pic]

And the direction of acceleration a, and F are the same and so the centripetal force acts towards the centre of the circle.

[pic]

Note if an object is moving around a circle the angle at the centre of the circle is also changing. The symbol for angle is θ and the S.I. unit in which it is measured is radians.

Angular velocity is defined therefore as the rate of change of angle symbol ω units radians per second.

And remember that

[pic]

[pic]

[pic]

This is the equation for the magnitude of the centripetal force and the direction is in the same direction as ac i.e. towards the centre of the circle.

Example a body of mass 2 kg moves around a circular path of radius 50 cm at a constant speed of 10 m.s-1 .

Calculate (i) the angular speed and

(ii) the centripetal force.

For circular motion we have shown that

[pic]

It can also be shown that

[pic]

There is also an equivalent set of equations of motion in terms of the angular values

[pic]

It is not the case that ωf = v except in the case where the radius of the circle is equal to 1m.

PROBLEM SHEET.

Question 1. A car which has wheels of radius 30 cm starts from rest and accelerates uniformly to a speed of 15 m.s-1 in a time of 8seconds. Calculate (a) angular acceleration (b)time taken to rotate 100 times.

Question 2. A spin drier revolving at 900r.p.m.( revolutions per minute ) slows down uniformly to 300 r.p.m. while making 50 revolutions.

Calculate (a) angular acceleration and (b) time taken

Question 3. In the hydrogen atom an electron rotates around a circle of radius

0.5 x 10-10 m. The centripetal force being the attraction of the -ve electron to the +ve nucleus. Calculate the magnitude of the force if the speed of the electron is 2.3 x 106 m.s-1 and the mass of the electron is

9 x 10-31 kg.

Newton’s Second Law

Used to derive the equation for Force,

It say that the force is proportional to the rate of change of momentum

i.e.

[pic]

However

[pic]

Where k is a constant.

It was found by experimental techniques that the constant k is equal to 1

Therefore

[pic]

NEWTON’S THIRD LAW.

Two bodies in contact.

e.g A block of mass m ( kg ) placed on a level surface.

F2

Mass m (kg )

F1

F1 is the weight of the mass m (kg ) acting in the downward vertical direction.

F1 = m . g (Action Force) = W

F1 is the force on the second body i.e. the level surface due to the first body i.e. the mass

By Newton’s Third Law the second body, the plane surface must exert an equal and opposite force on the mass shown as F2

Therefore

F2 = - F1 = - m . g = R

Called the Reaction force, symbol R

The total force on the mass in the vertical direction is equal to the sum of the forces acting in the vertical direction

There are two forces acting in the vertical direction

So

FT = F1 + F2 = m . g + (- m . g) = 0

This is correct since the mass is stationary in the vertical direction the total force in that direction should equal zero.

NOTES:

1. ( F1 ( = ( F2 (

i.e. the magnitude of each of the forces is the same.

2. F1 and F2 are in opposite directions. In vector notation we use the negative to indicate opposite direction. So if we use +ve for the vertically down direction then the vertically upward direction is given a –ve value.

3. It is also important to remember that the Reaction force between two bodies always takes place in the direction perpendicular to the direction of the common surface of the two bodies.

INCLINED PLANE.

Consider now a mass m (kg) placed on a plane inclined at θo to the horizontal or + x direction

“n object of mass m ( kg ) is placed on an inclined plane.

The force acting on the object is its weight which acts vertically downward

m(kg) F = W = m .g θ

i.e. [pic]

To calculate a value for the force tending to pull the mass down the inclined plane we must find the component of F which acts in that direction i.e. the component of F which acts along ( or parallel to ) the incline.

We need therefore to resolve F into two components

F1 which is parallel to the incline plane and

F2 which is perpendicular to the incline plane.

Remember once the direction of F1 is chosen the direction of F2 must be perpendicular to it.

Check resolving vectors into components.

F1 F2

F W θ

Το calculate the component of F = W = m .g we must

1. To calculate F1 drop a perpendicular from F to the direction parallel to the incline plane.

2. To calculate F2 drop a perpendicular from F to the direction perpendicular to the incline

3. It is also interesting to note that the perpendicular that is dropped from F to the direction parallel to the incline, to give the vector F1, is equal to F2. And the perpendicular dropped to the direction perpendicular to the incline to calculate F2 is equal to F1

Notice that if θ is the angle between the incline and the + x direction then the angle between F and the direction perpendicular to the incline is also θ.

Looking at the right angled triangle made by the vectors F , F1 and the vector equal to F2 we get

[pic]

[pic]

Remember θ is the angle between the incline and the +_x direction.

Again be Newtons third law there is also a reaction force which takes place in the direction perpendicular to the incline i.e.

[pic]

NOTES :

1. F1 and F2 are the components of F in two directions parallel to and perpendicular to the incline and

F1 + F2 = F

2.

R mgSinθ mgCosθ

The total force on an object in any direction is equal to the sum of the forces acting on the object in that direction.

The total force on the mass therefore in the direction parallel to the incline is equal to

[pic]

Therefore the acceleration in that direction may be calculated

[pic]

Therefore there is an acceleration and so the object moves down the incline.

The total force in the direction perpendicular to the incline is equal to the sum of the forces in that direction i.e. the sum of the forces F2 and R

[pic]

Correct since the object ( mass ) is NOT moving in the direction perpendicular to the incline.

[pic]

So for a level surface

[pic]

and since F1 = 0 the object does not move along the level surface ALL CORRECT

PROBLEM SHEET

Question 1. A block of mass 10 kg rests on a level surface. Calculate (i) the weight of the block (ii) the reaction force on the block due to the level surface.

Question 2. A block of mass 5 kg is placed at rest on an inclined plane, where the angle between the incline and the +x axis is 50o.

(a)Draw a vector diagram to represent all the forces acting on the block .

b) calculate the acceleration of the block down the incline

c) the force parallel to the incline

d) the reaction force on the block

e) how long will it take the block to slide 0.25 m down the incline.

Question 3. A mass of 10kg takes a time of 0.5 s to travel 61.31 cm down an inclined plane. Assuming the initial velocity is zero calculate (a) the angle θ (b) F1 (c) F2 and (d) R

FRICTION.

Friction is defined as the resistance force which must be overcome when one surface moves over another surface.

Friction is a force with S.I. Units Newtons and it is vector whose direction is always opposite to movement.

Friction therefore always opposes motion.

Consider again a block of mass m ( kg) resting on a level surface.

An applied force acts on the block in the horizontal direction

R

FAPP Ff

W

There are now four forces acting on the block

i) its weight acting vertically down

ii) the reaction force, R = - W

iii) the applied force, in the +x direction and

iv) the frictional force, in the – x direction.

It can be shown that

[pic]

i.e. the value for the magnitude of Ff

Where μ is the coefficient of friction between the two surfaces and R is the reaction force.

NOTE:

1. Ff is the minimum applied force which will move the block and is in the opposite direction to the applied force

2. the Reaction force and it is in the direction perpendicular to the plane common to the two surfaces

3. The coefficient of friction μ is a scalar quantity between 0 and 1 which is determined by the nature of the two surfaces.

4. Ff is a vector quantity which means that we need to specify both magnitude and direction the magnitude is given by the equation

[pic]

and the direction always opposes motion.

On a level surface the total force, in the horizontal direction, is equal to the sum of the forces in the horizontal direction

[pic]

The total force in the horizontal direction =

Mass of object X acceleration in the horizontal direction

[pic]

Total force in the vertical direction = sum of the forces in the vertical direction.

[pic]

Correct since there is no movement in the vertical direction.

FRICTION AND THE INCLINED PLANE

R

F1 F2

Ff W

When a mass m ( kg) is placed on the incline plane with friction there is a frictional force on the mass opposing motion i.e up the incline

There are two forces parallel to the incline plane

F1 down the incline and Ff up the incline

The total force parallel to the incline is equal to the sum of the two forces along the incline i.e. F1 + Ff

By Newton’s Third Law

[pic]

Also we know that

[pic]

There the total force parallel to the incline.

[pic]

[pic]

Remember that to calculate the acceleration in any direction divide the total force in that direction by the mass.

The general equation are

[pic]

For a level surface with no friction

μ ’ 0 and θ ’ 0

Therefore the equations become

[pic]

Incline with no friction μ ’ 0 and

angle = θ

[pic]

Level surface with friction θ ’ 0 and

coefficient of friction = μ

[pic]

PROBLEM SHEET.

Question 1. A wooden block of mass

3 kg rests on a level surface where the coefficient of friction between the block and the surface is 0.4. Calculate (i) the minimum force required to move the block

(ii) If a force of 60 N is applied to the block in the horizontal direction calculate the acceleration on the block.

Question 2. A block of mass 6 kg is place on an inclined plane where the angle between the incline and + x direction is 35o. Draw a vector diagram to show all the forces acting. Calculate

i) the force on the block down the incline

ii) the acceleration if the plane is frictionless

iii) the acceleration is the coefficient of friction is 0.2

iv) If an applied force of 50 N acts on the mass down the incline repeat the above calculations.

Question 3.(a) How long will it take a wooden block to slide from rest a distance of 2 m down a slope inclined at 30o to the horizontal if the coefficient of friction between the block and the surface is 0.25 and the mass of the block is 10 kg.

b) Calculate the time if the plane is frictionless.

c) Calculate the time if the coefficient of friction is 0.4.

Question 4. A ball rolls down an inclined plane a distance of 4 m in 10 seconds. Calculate

i) the acceleration down the incline

(ii) if the incline is at 40o to the + x direction calculate the coefficient of friction between the incline and the ball.

Question 5. A mass of 80 kg is pulled along a level road by a rope which is inclined at 30o to the horizontal. The force applied to the rope is 150 N.

Calculate (i) the distance travelled along the road in 1 minute assuming zero friction and initial velocity zero.

(ii) If the coefficient of friction between the mass and the road is 0.2 calculate the distance travelled.

Work energy and power.

Definitions:

Work is defined as the product of the force applied by the distance moved in the direction of the force.

Energy is defined as the ability to do work

Power is defined as the rate of doing work

Work

We get the following equation from the definition

[pic]

Called the Joule Symbol J

One Joule of work is done when a force on one Newton is moved through a distance of one metre measured in the direction of the force.

Work done = Energy acquired

Unit of energy and the unit of work are the same therefore the S.I. unit of energy is the Joule

Conservation of Energy.

Conservation of energy states that energy can neither be created nor destroyed only converted from one form to another.

The total energy in a closed system is constant.

Potential Energy.

Potential Energy of a body is the energy it possesses due to its position.

Example is the gravitational potential energy stored in a body at a height of h (m) above the ground or zero of potential energy.

Recall that work done = energy acquired

Therefore the work done in moving the body from the ground ( taking ground as the zero of potential energy ) to the height h (m) is the energy of the body at that point.

[pic]

Kinetic Energy.

The Kinetic energy of a body is the energy it possesses due to its motion.

When a force acting on a body produces movement the force does work and the body possesses kinetic energy because of its motion.

Again work done = energy acquired

[pic]

If we assume that for the motion the body started at an initial velocity of zero then u = 0

[pic]

Therefore the Kinetic energy of a body moving at a velocity v ( m.s-1) is given by

[pic]

Note : The zero of P.E is taken as the lowest point of the system and the zero of K.E. is taken when the body is stationary.

Total Mechanical Energy equals the sum of the potential and the kinetic energy of the body and for a closed system

Total energy = P.E. + K.E. = Constant

Conservation of Energy example

Consider a body of mass m ( kg) which is at a height h(m) above ground and is allowed to free-fall from the height to ground.

We will consider the body at three positions

1. At the maximum height just before it is dropped

2. At the half way position

3. At the ground level just before it impacts with the ground

At position 1. The body is stationary

[pic]

At position 2 , the body has dropped a distance of h/2 metres, under acceleration due to gravity, from the top and is at a height of h/2 metres above ground.

[pic]

At position 3 just before impact with the ground the body has the body has dropped a distance of h metres, under acceleration due to gravity, from the top and is at a ground level

[pic]

Therefore the total energy of the system remains constant.

Example 1. A body of mass 5 kg is dropped from the top of a building 100 m high.

Calculate

i) The total energy of the body before it is dropped

ii) The kinetic energy at 70m from the ground

iii) The total energy at 70 m from the ground

iv) The velocity at 40m from the ground

v) The kinetic energy at 40m from the ground

vi) The total energy at 40m from the ground

vii) The kinetic energy just before impact with the ground

viii) Velocity just before impact with the ground

POWER Defined as the rate of doing work.

[pic]

Efficiency.

Simple and complex machines that do work have in mechanical parts that move, so some energy is always lost because of friction, heat, sound etc.

Thus not all of the input energy goes into doing useful work

However remember the total energy is not lost either it is in the heat light sound energy.

Total Input energy = Total output energy

Total Input = Useful output + Heat , light etc

Mechanical Efficiency is essentially a measure of what you get out of a machine that is useful for the amount of input expressed as a fraction or a percentage.

Efficiency E = Work output x 100

Energy input

S .I. Units NONE it is a % value.

Given that

[pic]

The equation can also be written as

[pic]

Example 1. An electric pump raises 8,950kg of water from a reservoir whose storage level is 4 m below the surface to a storage tank 31 m above ground. If the operation takes one hour to complete find the power rating of the pump if it is 70% efficient.

Question 2. An engine used to pump water out of a shaft 50 m deep discharges it at the surface with a speed of 20 m.s-1.

If 2 kg of water is removed every second and the engine is 80% efficient what is the power rating.

Question 3. A 1200 kg car can accelerate from rest to a speed of 25 m.s-1 in 8 s. Calculate the power of the engine.

Question 4. What power does a person expend on a 100 kg object being dragged down a slope of angle 20o to the horizontal with a speed of 0.5 m.s-1 if the coefficient of friction is 0.9

Question 5. A simple pendulum of length

1m is oscillating 30o each side of the normal. What is the velocity of the pendulum bob at its lowest point.

Question 6. A 400 g mass hangs from the end of a string of length 50 cm. A bullet of mass 50 g travelling horizontally hits the mass and sticks to it. The system is deflected through an angle of 40o. Calculate the velocity of the bullet just before impact.

Question 7. A crane of power rating 65 kW can lift a 2000 kg mass to a height of 12 m in 8 s. Calculate the efficiency of the crane.

Question 8. Calculate the power rating of an engine which can accelerate a 50 kg object from rest to 12 km.hr-1 in 5 s along a frictionless level surface

Question 9. Repeat question 8 if the coefficient of friction between the object and the plane is 0.4

The Law of Conservation of Energy 2

Energy in a system may take on various forms (e.g. kinetic, potential, heat, light). The law of conservation of energy states that energy may neither be created nor destroyed. Therefore the sum of all the energies in the system is a constant.

The most commonly used example is the pendulum:

Position 1 and 3 maximum positions

Position 2 middle position and lowest point of the system

The formula to calculate the potential energy is:

PE = mgh

The mass of the ball = 10kg 


height, h = 0.2m 


acceleration due to gravity g = 9.8 m/s2


Substitute the values into the formula and you get:

PE = 19.6J (J = Joules, unit of energy)

At position 1 is the Potential Energy (PE) = 19.6J while the Kinetic Energy (KE) = 0.

As the blue ball is approaching position 2 the PE is decreasing while the KE is increasing. At exactly halfway between the two positions PE = KE.

At position 2 the Kinetic Energy is at its maximum while the Potential Energy (PE) = 0.

At this point, theoretically, all the PE has transformed into KE.

Therefore now the KE = 19.6J while the PE = 0.

At position 3 Potential Energy (PE) is once again at its maximum and the Kinetic Energy (KE) = 0.

Total Mechanical Energy = PE + KE

NOTE: This is with the abscence of outside forces such as friction

Using our common sense we know that it's impossible for the pendulum to swing higher than the height h without giving it a push yourself. If there was no friction, the pendulum would swing back and forth forever because of the law of conservation of energy.

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