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PC1221 Fundamentals of Physics I

Lectures 13 and 14

Energy and Energy Transfer

Dr Tay Seng Chuan

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Ground RulesGround Rules

Switch off your handphone and pagerSwitch off your laptop computer and keep itNo talking while lecture is going onNo gossiping while the lecture is going onRaise your hand if you have question to askBe on time for lectureBe on time to come back from the recess break to continue the lectureBring your lecturenotes to lecture

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1. If vectors A and B (collinear) pointing in the same direction are added, the resultant has a magnitude equal to 4.0. If B is subtracted from A, the resultant has a magnitude equal to 8.0. What is the magnitude of A? A. 2.0B. 3.0C. 4.0D. 5.0E. 6.0

Answer:A and B are parallel.|A| + |B| = 4 …(1)|A| - |B| = 8 …(2)(1) + (2) 2|A| = 12, so |A|= 6.

(If A and B are opposite, |A| can be |-2|, and |B| can be |6|.)

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2. The site from which an airplane takes off is the origin. The x-axis points east; the y-axis points straight up. The position and velocity vectors of the plane at a later time are given by.

The plane is most likelya. just touching down.b. in level flight in the air.c. ascending.d. descending.e. taking off.

sm 224 and m )1014.9ˆ1061.1( 36 ivjir +=×+×=

Answer:The y value is not equal to 0 meaning the airplane is above the ground. The airplane is cruising toward east at an altitude of about 9 km.

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3. If P = 6.0 N, what is the magnitude of the force exerted on block 1 by block 2?

A. 6.4 NB. 5.6 NC. 4.8 ND. 7.2 NE. 8.4 N

Answer:

Let F1 be the force in between block 1 and block 2, and F2 be the force in between Block 2 and Block 3. Let a be the acceleration.

6 - F1 = 2a …(1)F1 - F2= 3a …(2)F2 = 5a …(3) Substitute (3) in (2), F1-5a= 3a, i.e., F1= 8a …(4)Substitute (4) in (1), 6 - 8a=2a, i.e., 6 = 10a,

so a= 0.6 m/s2. Using (1), F1 = 6 - 2a = 6 - 2x0.6 = 6 - 1.2 = 4.8N

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P2.0 kg2.0 kg2.0 kg

3.0 kg3.0 kg3.0 kg5.0 kg5.0 kg5.0 kg

F1 F1 F2 F2

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4. A roller-coaster car has a mass of 500 kg when fully loaded with passengers. The car passes over a hill of radius 15 m, as shown. At the top of the hill, the car has a speed of 8.0 m/s. What is the force of the track on the car at the top of the hill?

A. 7.0 kN upB. 7.0 kN downC. 2.8 kN downD. 2.8 kN upE. 5.6 kN down

Answer:mg – R = m v2/r, so

R = mg - m v2/r = 500x9.8 – 500 x 82/15 = 2766.66 N = 2.8 kN up

15 m

8.0 m/s

15 m

mg

R

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SM2 StudentsSM2 Students are to attend a are to attend a seminar this Thursday at 2pm, seminar this Thursday at 2pm, LT32. Attendance is LT32. Attendance is compulsory. compulsory.

Class monitor is to take Class monitor is to take attendance at 1:40pm.attendance at 1:40pm.

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Ground RulesGround Rules

Switch off your handphone and pagerSwitch off your laptop computer and keep itNo talking while lecture is going onNo gossiping while the lecture is going onRaise your hand if you have question to askBe on time for lectureBe on time to come back from the recess break to continue the lectureBring your lecturenotes to lecture

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Introduction to Energy

The concept of energy is one of the most important topics in science while it is not the only important topicEvery physical process that occurs in the Universe involves energy and energy transfers or energy transformationsGiving a lecture now is an example of the transfer and transform of energy but it is more complicated (chemical energy from food I ate this morning to the sound energy in my voice now)

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Energy Approach to Problems

The energy approach to describing motion is particularly useful when the force is not constantAn approach will involve Conservation of Energy

This could be extended to biological organisms, technological systems and engineering situations

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SystemsA system is a small portion of the UniverseA valid system may

be a single object or particlebe a collection of objects or particlesbe a region of spacevary in size and shape

Energy in a system is conserved.

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WorkWork

The work, W, done on a system by an agent exerting a constant force on the system is the product of the magnitude, F, of the force, the magnitude Δr of the displacement of the point of application of the force, and cos θ, where θ is the angle between the force and the displacement vectors. W = F Δr cos θ

θF

F cosθΔr

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Work, cont.

W = F Δr cos θThe displacement is that of the point of application of the forceA force does no work on the object if the force does not move through a displacementThe work done by a force on a moving object is zero when the force applied is perpendicular to the displacement of its point of application

θF

F cosθΔr

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Did you do any work if you were repelling along a rope down a helicopter?

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θF

F cosθΔr

Did you do any work if you were running on a level ground?

Why did you feel tired after the run?

Take up Physics major Take up Physics major –– biophysics. This biophysics. This course will give you interesting knowledge.course will give you interesting knowledge.

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Example. Batman, whose mass is 80.0 kg, is dangling on the free end of a 12.0-m rope, the other end of which is fixed to a tree limb above. He is able to get the rope in motion as only Batman knows how, eventually getting it to swing enough that he can reach a ledge when the rope makes a 60.0° angle with the vertical. How much work was done by the gravitational force on Batman in this maneuver?

Answer:

The force of gravity on Batman is down. Only his vertical displacement contributes to the work gravity does. His original y-coordinate below the tree limb is -12 m. His change in elevation (the distance traveled) isΔy = -12 cos (60°) m – (-12 m) = 6 m.

The work done by gravity is80 kg x 9.8 m/s2 x 6 m = 4704 J

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Work in Pushing a BlockThe normal force, n, and the gravitational force, m g, do no work on the object

cos θ = cos 90° = 0

The force F does do work on the object

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More About WorkThe system and the environment must be determined when dealing with work

The environment does work on the systemWork by the environment on the system

The sign of the work depends on the direction of F relative to Δr

Work is positive when projection of F onto Δr is in the same direction as the displacementWork is negative when the projection is in the opposite direction

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Units of Work

Work is a scalar quantityThe unit of work is a joule (J)

1 joule = 1 newton . 1 meterJ = N · m

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Work Is An Energy TransferIf the work is done on a system and it is positive, energy is transferred to the systemIf the work done on the system is negative, energy is transferred from the systemIf a system interacts with its environment, this interaction can be described as a transfer of energy across the system boundary (e.g., heat generated due to friction)

This will result in a change in the amount of energy stored in the system

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Scalar Product of Two VectorsThe scalar product of two vectors is written as A . B(read as A dot B)

It is also called the dot product

A . B = A B cos θθ is the angle between A and B

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Scalar Product, cont

The scalar product is commutativeA . B = B . A

The scalar product obeys the distributive law of multiplicationA . (B + C) = A . B + A . C

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Dot Products of Unit Vectors

Using component form with A and B:

0kjkiji

1kkjjii

=⋅=⋅=⋅

=⋅=⋅=⋅

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

++=⋅

++=

++=

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Work Done by a Varying Force

Assume that during a very small displacement, Δx, Fis constantFor that displacement, W~ F ΔxFor all of the intervals,

f

i

x

xx

W F x≈ Δ∑

Force

Displacement

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Work Done by a Varying Force, cont

Therefore,

The work done is equal to the area under the curve of force versus displacement

lim0

ff

ii

xx

x x xxx

F x F dxΔ → Δ =∑ ∫f

i

x

xxW F dx= ∫

Force

Displacement

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Work Done By Multiple Forces

If more than one force acts on a system and the system can be modeled as a particle, the total work done on the system is the work done by the net force

( )f

i

x

net xxW W F dx= =∑ ∑∫

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Work Done by Multiple Forces, cont.

If the system cannot be modeled as a particle, then the total work is equal to the algebraic sum of the work done by the individual forces

net by individual forcesW W= ∑

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Hooke’s Law

The force exerted by the spring is Fs = - kx

x is the position of the block with respect to the equilibrium position (x = 0)k is called the spring constant or force constant and measures the stiffness of the spring

This is called HookeHooke’’s Laws Law

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Hooke’s Law, cont.When x is positive (spring is stretched), F is negativeWhen x is 0 (at the equilibrium position), F is 0When x is negative (spring is compressed), F is positive

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Hooke’s Law, finalThe force exerted by the spring is always directed opposite to the displacement from equilibriumF is called the restoring forceIf the block is released it will oscillate back and forth between x and –x

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Work Done by a SpringIdentify the block as the systemCalculate the work as the block moves from xi = - xmax to xf = 0, or

xi = xmax to xf = 0

The total work done as the block moves from–xmax to xmax is zero

( )max

0 2max

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f

i

x

s xx xW F dx kx dx kx

−= = − =∫ ∫

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Spring with an Applied ForceSuppose an external agent, Fapp, stretches the springThe applied force is equal and opposite to the spring forceFapp = -Fs = -(-kx) = kxWork done by Fapp is equal to ½ kx2

max

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Example. If it takes 4.00 J of work to stretch a Hooke's-law spring 10.0 cm from its unstressed length, determine the extra work required to stretch it an additional 10.0 cm. Answer:

To stretch the spring to 0.200 m requires

( )214.00 J 0.100 m2k=

800 N mk∴ =

( ) ( )21 800 0.200 4.00 J 12.0 J2WΔ = − =

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Kinetic Energy (F x)Kinetic Energy (F x)

Kinetic Energy is the energy of a particle due to its motion

K = ½ mv2

K is the kinetic energym is the mass of the particlev is the speed of the particlethe expression ½ mv2 can be derived using F = ma, and

A change in kinetic energy is one possible result of doing work to transfer energy into a system

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Kinetic Energy, contCalculating the workby integration:

2 21 12 2

f f

i i

f

i

x x

x x

v

v

f i

W F dx ma dx

W mv dv

W mv mv

= =

=

= −

∑∫ ∫

∫

∑ ma dx = m dx = mv dvdtdv

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Work-Kinetic Energy TheoremThe Work-Kinetic Energy Principle states ΣW = Kf – Ki = ΔKIn the case in which work is done on a system and the only change in the system is in its speed, the work done by the net force equals the change in kinetic energy of the system.Kinetic energy possessed by an object of mass m and velocity v isK = ½ mv2

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Work-Kinetic Energy Theorem

The normal and gravitational forces do no work since they are perpendicular to the direction of the displacementW = F Δx = ΔK

= ½ mv2 – 0(the vertical normal force has no effect)

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Nonisolated SystemA nonisolated system is one that interacts with or is influenced by its environment, e.g., push a block on a rough surface

An isolated system would not interact with its environment (e.g., push a block on a friction-less surface.)

The Work-Kinetic Energy Theorem can also beapplied to nonisolated systems. In that case there will be a transfer of energy across the boundary of an object (e.g., from the block to the surface where heat is generated.)

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Internal Energy

The energy associated with an object’s temperature is called its internal energy, Eint

In this example, the surface is the systemThe friction does work and increases the internal energy of the surface

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Potential EnergyPotential energy is energy related to the configuration of a system in which the components of the system interact by forcesExamples include:

elastic potential energy – stored in a springgravitational potential energy, e.g., you stand on top of Bukit Timah Hillelectrical potential energy, e.g., you switch on electricity to iron your shirt

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Example. A 2.00-kg block is attached to a spring of force constant 500 N/m as shown in figure. The block is pulled 5.00 cm to the right of equilibrium and released from rest. Find thespeed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the coefficient of frictionbetween block and surface is 0.350.

Answer:

(a) When the block is pulled back to equilibrium position, the potential energy in spring will be fully converted to the kinetic energy in the block when the spring returns to original length, i.e.,

½ k x2 = ½ m v2.

So, v = x x = x 0.05 = 0.79 m/smk

2500

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(b) If there is fiction on the surface, the potential energy on the spring is first wasted on the work done by friction. The remaining energy is converted to the kinetic energy in the block, i.e., ½ k x2 - μmg x = ½ m v2.

In terms of initial sum and final sum of energies, you can also treat it as

½ k x2 = μmg x + ½ m v2

½ x 500 x 0.052 = 0.35 x 2 x 9.8 + ½ x 2 x v2

v = 0.53 m/s

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Example. The ball launcher in a pinball machine has a spring that has a force constant of 1.20 N/cm. The surface on which theball moves is inclined 10.0° with respect to the horizontal. If the spring is initially compressed 5.00 cm, find the launching speed of a 100-g ball when the plunger is released. Friction and the mass of the plunger are negligible.

Answer:

Initial energy stored in the spring

= ½ k x2 = ½ x 1.20 N/(10-2m) x (5 cm x 10-2)2 = 0.15 J

This 0.15 J is used to (i) move the ball up the slope of 10°for a vertical distance of 5 cm x sin (10 °) = 0.87 cm, and (ii) provide the ball with a muzzle speed of v. So we have

0.1 kg x 9.8 m/s2 x 0.0087 m + ½ x 0.1 kg x v2 = 0.15 J

v = 1.68 m/s44

When pushing the crate with a force parallel to the ground, the force of friction acting to impede its motion is proportional to the normal force acting on the crate—in this situation, the normal force is equal to the crate’s weight. When pulling the crate with a rope angled above the horizontal, the normal force on the crate is less than its weight—the force of friction is therefore reduced. To keep the crate moving across the floor, the applied force in the parallel direction must be greater than or equal to the force of friction—pulling on the rope therefore requires a smaller parallel applied force. The work done in moving an object is equal to the product of the displacement through which it has been moved and the force component parallel to the direction of motion. The applied force component parallel to the ground is smaller when pulling the crate with the rope—thus, the work done to move the crate with the rope must be less, regardless of the weight of the crate or the displacement.

Example. You need to move a heavy crate by sliding it across a flat floor with a coefficient of sliding friction of 0.2. You can either push the crate horizontally or pull the crate using an attached rope. When you pull on the rope, it makes 30° angle with the floor. Which way should you choose to move the crate so that you will do the least amount of work? How can you answer this question without knowing the weight of the crate or the displacement of the crate?

Answer:

Normal forceNormal

force

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F2 cos 30° = µ (mg – F2 sin 30°)

F2 cos 30° + µ F2 sin 30° = µ mg

F2 (0.866 + 0.5) = µ mg

F2 = 0.732 µmg = 0.732 F1

Work done = F2 x d

= 0.732 F1 x d

F1 = µmg

Work done = F1 x d

Normal force

Normal force

F1F2

mg mg

PushPush PullPull

What if you pull at 0 What if you pull at 0 °°, i.e., in , i.e., in parallel with the surface?parallel with the surface?

µ = 1 µ = 1

s1Slide 45

s1 scitaysc, 10/2/2007

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Ways to Transfer Energy Into or Out of A System

Work – transfer by applying a force and causing a displacement of the point of application of the force, e.g., you push a bookMechanical Waves – allow a disturbance to propagate through a medium, e.g., sound, earthquake, tsunamiHeat – is driven by a temperature difference between two regions in space, e.g., hot tea

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More Ways to Transfer Energy Into or Out of A System

Matter Transfer – matter physically crosses the boundary of the system, carrying energy with it, e.g., use of gasoline in carElectrical Transmission – transfer is by electric current, e.g., use of electrical hair dryerElectromagnetic Radiation –energy is transferred by electromagnetic waves, e.g., use of light bulb.

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Conservation of Energy

Energy is conservedThis means that energy cannot be created or destroyedIf the total amount of energy in a system changes, it can only be due to the fact that energy has crossed the boundary of the system by some method of energy transfer

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PowerPower

The time rate of energy transfer is called powerThe average power is given by

when the method of energy transfer is work

WPt

=Δ

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Instantaneous Power

The instantaneous power is the limiting value of the average power as Δt approaches zero

This can also be written as

lim0t

W dWPt dtΔ →= =

Δ

dW drP F F vdt dt

= = ⋅ = ⋅

Power = force x velocity51

Units of PowerThe SI unit of power is called the watt

1 watt = 1 joule / second = ((kg x m/s2) x m ) / s = 1 kg . m2 / s3

A unit of power in the US Customary system is horsepower

1 hp = 746 WUnits of power can also be used to express units of work or energy

1 kWh = (1000 W)(3600 s) = 3.6 x106 JEnergy transferred in 1 hr at a constant rateNot 1 kilo-watt

per hour

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Example. An energy-efficient light bulb, taking in 28.0 W of power, can produce the same level of brightness as a conventional bulb operating at power 100 W.

The lifetime of the energy efficient bulb is 10 000 h and its purchase price is $17.0, whereas the conventional bulb has lifetime 750 h and costs $0.420 per bulb. Determine the total savings obtained by using one energy-efficient bulb over its lifetime, as opposed to using conventional bulbs over the same time

period. Assume an energy cost of $0.080 per kilowatt-hour.Answer:

kWh

kWh

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Energy and the Automobile

The concepts of energy, power, and friction help to analyze automobile fuel consumptionAbout 67% of the energy available from the fuel is lost in the engineAbout 10% is lost due to friction in the transmission, drive shaft, bearings, etc.

About 6% goes to internal energy and 4% to operate the fuel and oil pumps and accessories

This leaves about 13% to actually propel the car

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Friction in a Car

The magnitude of the total friction force (ƒt ) is the sum of the rolling friction (ƒr ) force and air resistance (ƒa)

ƒt = ƒr + ƒa

At low speeds, rolling friction predominatesAt high speeds, air drag predominates

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Friction in a Car, cont