Physics

By the end of the unit, the learner should be able to evaluate the relation between work, energy and power and the resulting phenomena.

By the end of this unit, I will be able to:

* define work done, energy and power.

* state the formulae of work, energy and power.

* explain how power depends on energy.

* explain how gravitational potential energy.

* identify the difference between potential energy and kinetic energy

* describe strain and work done in deforming materials

Introduction

In real life, we always use the term work. Which means “task to be accomplished. But before the task to be done, one must have energy. Then if a given work is done in a given time, we say that one has power i.e work done in a given time.

Work

Review of the idea of work

Activity 1

Study and interprete the diagram below

CASE I

Work is done when a force moves its point of application along the direction of its action.

* How is the force applied onto the body?

* Why does it change its position?

* What if the body is 10 times the mass of the boy. Would the body change its position? Why?

* State the direction of application of force.

From the fig. 4.1 and your deductions, how can you define Work?

CASE II

Aim; To relate distance,force and work

Let us as a class visit any where people are constructing a house,a bride, road.Ask them why they are paid?Ask them how they measure what they do.

From the diagram (CASE I)

The work is done when a force moves its point of application along the direction of its action.

Let W = Force × distance

W = F × d

1Nm=1Joule

In the second case, the work done is defined as the product of the component of the force in the direction of the motion and displacement in that direction.

That is: W = F ×COS θ

There is another unit of work called kilogram-metre, which is the work done by a force of 1kg when its application point moves through a distance of 1 m1 kgf = 9.81N

In cgs system, the work is expressed in (erg) 1erg = 10-7J, 1(erg)=1(dry n cm)

Then the Joule is the work done by a force of 1N when its application point moves through a distance of 1 meter in the direction of force.

Work is the scalar although force and displacement are both vectors.

Expressions of some kinds of work

Work of the gravitational force

Activity 3

a) Hold a book in your hands at a height say h.

b) Leave it to fall vertically onto the ground.

NOTE;

From the deductions, it can be noted that:

The work done by the gravitational force does not depend on the path followed but on the change of the height.

Work done by the force of pressure

Activity 4

Requirements

* A syringe with a piston.

Aim: To determine work done by a piston.

* Pull the piston through a small distance ∆x as shown in the figure below.

Assuming you applied a force F, What is the work done?

Note:

From Pressure being the force per unit area, we know that gas exerts pressure on the inside surface of the container.

Let us consider gas confined in a cylinder by a piston.If p is the pressure in the cylinder A, the area of the piston, heating the gas, the piston moves up, the volume increases but the pressure remains constant

.If ∆x is the displacement of the piston, the force which moves it is given by:

F = PxA The work during the displacement is:

W=Fx∆x

W=PxAx∆x

A x ∆x= ∆V, the change of the volume

Then we have: W=Px∆V =P(V2 - V1)

Energy

Ask yourself why some times you feel like not working or bored.

What do you normally say when you are asked why you are not performing any duty?

Use what comes into your mind to define energy.

Normally we say that Energy is ability of a body to do work.It’s measured in Joules like work.

When an interchange of energy occurs between two bodies, we can consider the work done as a measure of the quantity of energy transferred between them.

Potential energy

Quick Check 1

Use this principle to determine the blanks in the following diagram. Knowing that the potential energy at the top of the tall platform is 50 J, what is the potential energy at the other positions shown on the stair steps and the incline?

Activity 5

How do we know that things have energy just because of their height? Well, let’s think about the following process:

1. You lift a ball off the ground until it is above your head.

2. You drop it.

3. It is moving fast right before it hits the ground.

4. Draw a conclusion.

Energy is said to be conserved, which means that it cannot be created or destroyed, only transferred from one form into another. So whatever energy we put in, has to go somewhere.

Then we can define the potential energy as the energy a system of bodies has because of the relative position of its part .

Gravitational potential energy

From the diagram, what can you deduce?Still in a class, perform the same experiment using any mass.

Note:

* The potential energy when a body of mass m is at height h above ground level equals to the work which must be done against the downward pull of gravity to raise the body to this height.

* A force equal and opposite to W=mg has to be exerted on the body over displacement h assuming g is constant near the earth’s surface.

Therefore, work done by external force against gravity:

= force X displacement

= mgh

And so p.e = mgh

Activity 6

Aim ; To determine the effect of Gravitational potential energy

* The diagram.

* What do you think will happen when the car is made to slope down when its engine is switched off?

Check Your Understanding

Check your understanding of the concept of potential energy by answering the following questions.

1. A cart is loaded with bricks and pulled at constant speed along an inclined plane to the height of a seat-top. If the mass of the loaded cart is 3.0kg and the height of the seat top is 0.45 metres, then what is the potential energy of the loaded cart at the height of the seat-top?

Elastic potential energy

Individual Assignment

* Try to get a rubber band or any elastic material.

* Try to stretch as shown in the diagram.

* Explain what you have felt and seen.

Activity 7

Aim; To find out whether there is energy stored in elastic Materials

In laboratory,

* Try to perform experiment arranging your apparatus as shown in the figure below.

* What do you observe after putting a mass on the spring.

* What would happen if the mass of the body is given a small displacement downwards?

Kinetic energy

In all the diagrams indicated below, show the changes of energy of a moving object. study them clearly.

What energy do the objects you are seeing possess?

State the reason for your answer.

In general, we can define Kinetic energy as the energy a body has because of its motion.

Kinetic energy of an object in translational motion

Then if the centre of mass has a velocity v, the kinetic energy is given by:

Exercise

Study the graph below and answer the questions.

* Explain using equations where necessary how the values at different points were obtained.

* From the graph, what is the maximum energy of the system?

Kinetic energy theorem or work-energy theorem

We have seen that a moving object can do work. The opposite is true as well: work must be done on an object to give it k._e. To find the precise relationship, we reverse the above argument.

Suppose an object of mass m is moving with an initial speed v0 and to accelerate it (uniformly) to a speed v a net force is exerted on it parallel to its motion over a distance x.

Then the net work done on it is W = F × x

Using the second Newton’s law F = mγand we know that v² = v²0 + 2γx,

we find: W = F × x

Theorem: “The net work done on an object is equal to its change in kinetic energy”This is known as the work-energy theorem.

Total mechanical energy

The figure below is about a falling mango from a branch study it carefully.

* Are there energy changes?

* If yes, what are those energy changes?

We can therefore say that the total Mechanical energy is the sum of P. _e

and K._e. M._e = k._e + p._e

Conservation of mechanical energy

If a body of mass m is thrown vertically upwards with an initial velocity v₀ at A, it has to do work against the constant force of gravity.

Power

Activity 8

Interpret the diagrams below

* Have you ever done any of the two? If not, have you ever seen people doing such activities?

* If yes,how long did you take to accomplish the work?

From all the observations, power is the work done per unit time.

The concept power involved here is the rate of doing work. If an amount of work W is carried out a time t, the power for that time is defined to be:

We can also express the power delivered to a body in terms of the force that acts on the body and its velocity. Thus, for a particle moving in one dimension, the relation above becomes:

In the case of uniform motion, then the power delivered is p = Fv

Units

The S.I unit of power is the joule per second (J/S). This unit is used so often that it has been given a special name, the watt abbreviated (W) and 1(W) =(J/S).

Definition: A watt is the power when one joule of work is done for a second.

Linear momentum and impulse

Momentum

Momentum (symbol: p) of an object is the product of the mass and velocity of a moving body.

Note that since velocity is a vector quantity, momentum is also a vector quantity.

Momentum = mass × velocity,

Momentum depends on the mass of the object and its velocity. Momentum does not equal mass. (see Elephant)

Conservation of momentum

Activity 9: Field work

As a class, visit a place with a pool table.Let each and every body try to hit the ball using the playing stick.What happens when one ball hits another?

State and observe what you notice.

Draw a conclusion.

Activity 10: Fieldwork

* As a class, visit a place where there is billiard.

* Try to arrange the balls with the help of your teacher or any of the learners who have ever played it.

* Let one of you hit the white ball to strike/hit the rest.

* State what you observe after the white ball has hit the balls.

* Draw your conclusion from your observations.

* Repeat the same procedures using balls in the play grounds.

Suppose that a moving Ball A on the pool table of mass m₁ and velocitycollides with another ball B, of mass m₂ and velocity, moving in the same direction:

From Newton’s third law, the force F exerted by A on B is equal and opposite to that exerted by B on A. Also the time t during which the force acted on B is equal to the time during which the force of reaction acted on A.

Exercise

Truck Collision

Study the pictures below carefully

In a head-on collision:

Which truck will experience the greatest force?

Which truck will experience the greatest change in momentum?

Which truck will experience the greatest change in velocity?

Which truck will experience the greatest acceleration?

Which truck would you rather be in during the collision?

Impulse

Activity 11

* Move out the class to the play ground.

* In pairs (one pair at a time), kick a ball so that it is moving with a low speed. Let your friend stop it. Ask him/her what he/she felt. Let your friend do the same.

* For the second time, make sure that you kick the ball by applying a strong force so that it moves faster. Let your friend try to stop it. Ask him/her what this time he/she has felt?

* Go back in class and summarise

what you observed and felt while in the play ground.

Definition

If one exerts a force on moving a object in timet , the velocity of the object changes. We say that its momentum changes too.

The product of the force and the time t in which it acts is called impulse

represented by

In S.I units, the unit of impulse is Newton-second[Ns].

Exercise

A ball of mass of 0.4kg is thrown against a brick wall. It hits the wall moving horizontally to the left and rebounds to the right.

a) Find the Impulse of the net force on the ball during its collision with the wall.

b) If the ball is in contact with the wall for 0.01s, find the average force that the wall exerts on the ball during the collision.

Relationship between impulse and momentum

Suppose a force F acts on a body of mass m and gives it an acceleration a, the relationship between impulse and momentum can be seen by using Newton’s second law.

From Newton’s second law,

Applications

Slow down of a moving object by a constant force

Let us consider a rectilinear uniform motion of velocity v, of a moving object of mass m on which a retarding constant force acts parallel to the path. Let t be the time.

The impulse on the object is given by: Impulse = F. t

The final linear momentum is zero when the initial linear momentum is mv, its projection on the axis is given by:

Linear momentum = ⁺mv

Since the impulse is equal to change of momentum, it follows that Ft = 0 – mv, then

Recoil back of a rifle

Suppose a bullet of mass m, is, fired from a riffle of mass M with velocity . Initially the total linear momentum is zero. From the principle of the conservation of linear momentum, when the bullet is fired, the total momentum of bullet and rifle is still zero, since external force has acted on them. So if is the velocity of the rifle, we write:

WORK

Carefully interpret the diagram below.

Why do you think after firing the cap of soldier moved away from his head?Again, do you think that the soldier remained in same position as he was in before?

Explain your answer.

Exercise

1. Two bodies of mass 3kg and 5kg travelling in opposite directions on a horizontal surface collide. The velocities of the bodies before collision are 6m/s and 5m/s respectively. Given that after collision the two separate and move in the same direction in which the 5kg body was moving before collision, and the velocity of the 5kg mass is 1m/s, find the speed of the 3kg body after impact. Find also the loss in the energy.

2. A body of mass 2kg initially moving with a velocity of 1m/s is acted upon by a horizontal force of 6N for 3seconds. Find (i) Impulse given to the body (ii) Final speed of the body

Collisions

Activity 12

To know what happens to bodies after impact/colliding.

To know the effect of collision on the velocities and masses of bodies after colliding.

Observe the diagram below carefully assuming the black car to have a larger mass than a white one and answer the questions that follow.

Questions about the picture above:

a) Do you think after collision , the two cars continue moving? Explain why?

b) From what you observed, what is the effect of collision?

c) After separating the cars, do you think the masses of the cars changed? Explain why

In collision the total momentum of colliding objects is always conserved. Usually, however, their total kinetic energy is not conserved; some of it is changed to heat or sound energy, which is recoverable. Such collisions are said to be inelastic.

For example, when a lump of putty falls to the ground, the total momentum of putty and earth is conserved, that is, the putty loses momentum and the earth gains an equal amount of momentum. But all the kinetic energy of putty is changed to heat and sound on collision.

An Inelastic collision is the collision where the total kinetic energy is not conserved (total momentum always conserved in any type of collision). If the total kinetic energy is conserved, the collision is said to be elastic. For example, the collision between two smooth smoker balls is approximately elastic.

Elastic collision

In here, we shall consider objects colliding in a straight line and thereafter they move with different speeds in the same direction.

Elastic collision in one dimension

Let m₁ and m₂ be masses of two objects moving with speeds and .

Elastic collision in two or three dimensions

To understand this, use billiards as shown in the previous lesson

Some times after hitting the balls, they do not move in a straight line most especially (those who know how to play it) when you want to score in the centre hole. You must make sure that you hit the ball in target so that it moves at a certain angle.

Activity 13

• Try to hit a billiard ball as shown in figure above.

• Observe what happens when one ball hits another.

• Note down your observations.

• Present your findings/observation to the whole class.

On Striking the balls ,energy and momentum is conserved.

Conservation of momentum and energy can also be applied to collisions in two or three dimensions and in this case the vector nature of momentum is important.

One common type of non-head-on collision is one for which one particle (called the “projectile”) strikes a second particle initially at rest (the “tangent” particle). This is the common situation in games such as billiards.

The figure 4.30 shows particle 1 (the projectile m₁) heading along the x-axis towards particle 2 (the tangent m2) which is initially at rest. If these are, say, billiard balls, m₁ strikes m₂ and they go off at angles θ₁' and θ₂', which are measured relative to m₁’s initial direction (the x-axis)

Let us now apply the conservation of momentum and kinetic energy for an elastic collision like that on figure above. From conservation of kinetic energy, since v₂= 0, we have:

We choose the xy plane to be the plane in which lie the initial and final momenta. Since momentum is a vector, and is conserved, its components in x and y directions remain constant. In the x direction

Since there is no motion in the y direction initially, the y component of the total momentum is zero:

We have three independent equations. This means we can solve for utmost three unknowns. If we are given m₁, m₂, v (and v₂, if not zero), we cannot uniquely predict the final variables v'₁, v'₂, θ₁' and θ₂', since there are four of them, θ₂', for example, can be anything. However, if we measure one of these variables, say θ₁', then the other three variables (v'₁, v'₂, and θ₂') are uniquely determined and we can calculate them using the above three equations.

Inelastic collision

Collisions in which kinetic energies are not conserved are called inelastic collision. Some of the initial kinetic energy in such collision is transformed into other types of energy such as thermal or potential energy.

So the total final K._e is less than the total initial K._e, the inverse can also happen when potential energy (such as chemical or nuclear) is released and then the total final K.e can be greater than the initial K._e.

If two objects stick together as a result of a collision, the collision is said to be completely inelastic. Two colliding balls of putty that stick together or two railroad cars that couple together when they collide are examples.

On the diagram below, we have a collision in two dimensions of two vehicles on a road.

If the velocities of the two objects make a certain angle before collision and after collision they stick together, analytically we have this situation:

Division of mass

Let us consider a system constituted by two masses m₁ and m₂ which can slide without friction on a horizontal surface.

Initially, the two objects are linked and form a system. The centre of mass of the system has a uniform motion of velocity . The momentum is constant and equal to: p = (m₁ + m₂) vIf in the motion, there is division, the mass m₁ has the velocity v₁ and m₂ has the velocity v₂. The total momentum of the system is p¹= m₁v₁ +m₂v₂The conservation of momentum allows us to write:

p = p¹= (m₁ +m₂)v = m₁v₁ +m₂v₂

Examples

WORK, ENERGY AND POWER

1. Determine the work done by a horse exerting a force of 60kg on a vehicle when the vehicle travels a distance of 2 km

2. A 145-g baseball is thrown with a speed of 25m/s

a) What is its kinetic energy?

b) How much work was done to reach this speed starting from rest?

Student’s trials

1. A worker lifts up a stone of 3.5kg to a height of 1.80m each 30s. Find the work done in one hour.

2. Calculate the kinetic energy and the velocity required for a 70kg pole vaulter to pass over a 5.0m high bar. Assume the vaulter’s centre of mass is initially 0.90m off the ground and reaches its maximum height at the level of the bar itself.

3. Calculate the power required of a 1400kg car under the following circumstances

a) The car climbs a 10° hill at a steady 80km/h and

b) The car accelerates from 90 to 110km/h in 6.0s to pass another car on a level road. Assume the force of friction on the car is 700N in both parts of the problem.

4. A bullet is thrown obliquely in gravitational field, where g = 9.8 m/s2 with a speed of 20m/s. Calculate its speed when it reaches the height of 10m.5. A woman of mass 75kg walks up a mountain of height 20m.

a) What is the work done?

b) The walking up being done in 1.5 min, find the power,

c) What time will be taken by this woman to walk up the 20m in order to develop a power of 73W?

6. A stone of 2000kg falls from the top of a tower of height H = 200m. What is the total mechanical energy? What is the P. _e at height h =H/2 and its K._e?

7. Using the K._e. theorem, find the acceleration of the following system:

8. A small object A is suspended on a string of negligible mass of length OA=l making angleα with the vertical OB. One drops A without initial speed. Express, in function of l, g and α its speed when it passes in B.

9. A car travelling with a speed of 180km/hstrikes a wall. Find the height from which it will fall to produce the same energy. (g= 10m/s²).

10. An object of 2kg falls freely during 5s. What is the kinetic energy? What will be the kinetic energy if the object is thrown downward with the speed of 4m/s? g = 10m/s².

11. A small object A₀ of mass 50g is suspended by a string of 80cm of length of negligible mass. It’s moved away from the equilibrium position to the point A. The angle A₀OA being 60°, what is the change of the potential energy.

Linear momentum and impulse

1. What is the momentum of an 18.0g sparrow flying with a speed of 15.0m/s?

2. A moving object has an acceleration of 2.4m/s². It reaches in 12s a momentum of 800kgm/s. Compute the mass of that object and the force acting on it.

3. An object of mass 200g slides without friction on a horizontal surface and strikes a vertical obstacle and moves back following the same direction with a speed of 11m/s Find the impulse.

4. A system is constituted by two masses m₁=2kg and m₂=0.5kg connected by a string. The system moves on a horizontal table without friction from the rest. One makes it in motion applying an impulse of 10Ns but the string is cut. The result is, m₂ moves away with a certain speed and m1 with a speed of 2m/s What is the impulse received by m₁? by m₂? What is the speed of m₂ at the end of the impulse?

5. An object of mass m= 100g falls freely during 3s:

a) Find the received impulse,

b) Deduce the change of the speed.

c) Generalize to find the law of the free fall h = 12/gt²

6. One drops a ball of mass m from a height h₀ above the ground. The ball bounces till the point situated at the height h₁. Find the impulse received by the ball from the ground. Given that h₀= 2.55m, h₁= 2m, m= 0.2kg, g = 10m/s².

7. A tennis ball of mass 200g is thrown horizontally with a speed of 15m/s toward the north. Assuming that the ball and the racket are in contact during 0.01s, find the force that the player has to exert to return it back with a speed of 25m/s,

(a) toward the south,

(b) toward the south-east.

8. A 10,000kg railroad car travelling at a speed of 24.0m/s strikes an identical car at rest. If the cars lock together as result of collision, what is their common speed afterward?

9. Calculate the recoil back velocity of 4.0kg rifle which shoots a 0.050kg bullet at a speed of 280m/s

Extension

1. Suppose you throw a bowl of 0.4kg on a wall in bricks. It strikes the wall rolling horizontally leftward at 30m/s and rebounds horizontally rightward at 20m/s.

a) Find the impulse of the force exerted on the bowl by the wall.

b) If the bowl remains in contact with the wall during 0.01s, find the average force exerted on the bowl at the time of impact.

2. An automobile of mass m= 749.5kg accelerates from the rest. During the first ten seconds, the net force acting on it is given by the relation F = F₀ − kt, where F₀ = 888.6N, k = 44.48N/s and t is the time elapsed in second after the departure. Find the velocity at the end of the 10s and the travelled distance during that time.

3. A ball of mass 100g is dropped from a height h = 2m above the floor. It rebounds vertically to a height h' = 1.5m after colliding with the floor.

a) Find the momentum of the ball immediately before it collides with the floor and immediately after it rebounds.

b) Determine the average force exerted by the floor on the ball. Assume the time interval of the collision is 10^-2 s.

2. A proton travelling with a speed 8.2×105m/s collides elastically with a stationary proton in a hydrogen target. One of the protons is observed to be scattered at a 60° angle. At what angle will the second proton be observed, and what will be the velocities of the two protons after the collision?

3. A 15,000kg railroad car travels alone on a level frictionless track with a constant speed of 23.0m/s A 5000kg additional load is dropped onto the car. What then will be its speed?

4. A 90kg fullback is travelling 5.0m/s and is stopped by a tackler in 1s. Calculate

(a) the original momentum of the fullback,

(b) the impulse imparted to the tackler and

(c) the average force exerted on the tackler.

5. A billiard ball of mass m_A = 0.400kg moving with a speed v_A = 200m/s strikes a second ball, initially at rest, of mass m_B = 0.400kg. As a result of the collision, the first is deflected off at an angle of 30° with a speed of v_A = 1.20m/s .

(a)Taking the x to be the original direction of motion of ball A, write down the equations expressing the conservation of momentum for the components in the x and y directions separately,

(b) solve these equations for the speed, 'Bv, and angle α , of ball B. Assume the collision is elastic.

6. Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinates system. One is moving upward along the y axis at 3.00m/s, the other is moving to the right angle along the x axis with a speed of 4.80m/s After the collision (assumed elastic), the second ball is moving along the positive y axis. What is the final direction of the first ball, and what are their two speeds?

An explosion breaks a block of stone in three pieces A, B and C of respective masses m1, m2 and m3. Immediately after explosion the speeds v1=15m/s, v2=7m/s and v3=50m/s Vectors 1 and form a right angle. Assuming that m₁=1.5kg and m₂=3kg, determine the direction of and

Group work

1. Two masses m₁ = 5Kg and m₂ = 10Kg have velocities u₁ = 2m/s according to x positive axis and u₂ = 4m/s according to y positive axis. They collide and they get stuck. What is the final velocity after collision?

2. A lorry of transport of goods is empty and has a mass of 10000kg. When the lorry moves at 2m/s on a horizontal plane, it collides another lorry loaded, of total mass 20000kg; this last being initially at rest but with released breaks. If the two Lorries stuck together, after collision, what is their speed after collision?

3. a) With which velocity the loaded lorry must travel so that after collision the two remain at rest?

4. a) Distinguish between elastic collision and inelastic collision

b) Suppose two balls A and B of masses m₁ and m₂ are moving initially (in the same direction) along the same straight line with velocities u₁ and u₂ respectively. The two balls collide. Let the collision be perfect elastic. After collision, suppose v₁is the velocity of A and v₂ is the velocity of B along the same straight line. Prove that:

c) A ball of 0.1kg makes an elastic head-on collision with a ball of unknown mass that is initially at rest. If the 0.1kg ball rebounds at one third of its original speed, what is the mass of the other ball?

5. a) A 40g golf ball initially at rest is given a a speed of 30m/s when a club (a specially shaped stick for striking a ball) strikes. If the club and the ball are in contact for 1.5ms. what average force acts on the ball?

b) Is the effect of the ball’s weight during the time of contact significant? Why or why not?