Force

I.

Force

Why do objects move as they do? What makes an object at rest begin to move? What causes a body to accelerate or decelerate? What is involved when an object moves in a circle? Force. Intuitively, we experience force as any kind of a push or a pull on an object. When a motor lifts an elevator, or a hammer hits a nail, or the wind blows the leaves of a tree, a force is being exerted. We say that an object falls because of the force of gravity. Force, however, does not always give rise to motion, as when you push on a wall of a building! A force has magnitude as well as direction, and is indeed a vector that follows the rules of vector addition. We can represent any force on a diagram by an arrow just as we did with velocity. The direction of the arrow is the direction of the push or pull, and its length is drawn proportional to the magnitude of the force. II. Newton’s First Law of Motion Every body continues in its state of rest or of uniform speed in a straight line unless acted on by a nonzero net force. The tendency of a body to maintain its state of rest or of uniform motion in a straight line is called inertia. For this, Newton’s first law is often called the law of inertia.

III. Mass and Weight Mass is a measure of the inertia of a body. The more mass a body has, the harder it is to change its state of motion. It is harder to start it moving from rest, or to stop it when it is moving, or to change its motion sideways out of a straight-line path. A truck has much more inertia than a baseball, and it is much harder to speed it up or slow it down. It therefore has much more mass. In SI units, the standard unit of mass is the kilograms (kg.). The terms mass and weight are often confused with one another. Mass is a property of a body itself (it is a measure of a body’s inertia, or its “quantity of matter”).

Weight, on the other hand, is a force, the force of gravity acting on a body. Suppose we take an object to the Moon. The object will weigh only about one sixth as much as it did on Earth, since the force of gravity is weaker, but its mass will be the same.

IV. Newton’s Second Law of Motion What happens if a net force is exerted on a body? Newton’s first law perceived that the velocity will change. A net force exerted on an object may increase its speed, or, if in a direction opposite to the motion, it will reduce the speed. Since a change in speed or velocity is an acceleration, we can say that a net force gives rise to acceleration. The acceleration of a body is directly proportional to the net applied force. But the acceleration depends on the mass of the object as well. If you push an empty cart with the same force as you push one filled with goods, you will find that the latter accelerates more slowly. The greater the mass, the less the acceleration for the same net force. The mathematical relation is summarized by Newton’s Second Law of Motion: The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass, The direction of the acceleration is in the direction of the net force acting on the object. a = ∑F / m where a stands for acceleration, m for the mass, and ∑F for the net force. Rearranging, ∑F = ma The law defines force more precisely as an action capable of accelerating an object.

Units for Mass and Force System

Mass

Force

SI Units

Kilogram (Kg)

newton (N0

Definition Force required to accelerate 1 kg mass 1 m/s2

cgs

Gram (g)

dyne

Force required to accelerate 1 g mass 1 cm/s2

British

Slug

pound (lb)

Force required to accelerate 1 slug mass 1 ft/s2

Important: Use only one set of units in a given calculation or problem!

1 dyne = 10-5 N 1 pound ≈ 4.45 N Example. 1. Estimate the net force needed to accelerate a 1000 kg car at ½ g.

Given:

m = 1000 kg; a = ½ g = ½ (9.8 m/s2) ≈ 5 m/s2 ∑F = ma = (1000 kg) (5 m/s2) = 5000 N Ans.

2. What net force is required to bring a 1500-kg car to rest from a speed of 100

km/hr within a distance of 55 m? Given: m = 1500 kg; v1 =; 0 m/s; v0 = 100 km/hr = 28 m/s; x = 55 m.F = ? a) Find a:

v2 = v02 + 2ax a = - v02 / 2x = - 28 m/s ÷ 2 (55 m) = -7.11 m/s2 ∑F = ma = 1500 kg (-7.11 m/s2) = 1.1 x 104 N Answer [2.9]

Force Diagrams. To analyze forces and their effects on objects, the skill to use force diagrams is essential. There are two kinds:

System diagram: a sketch of all the objects involved in a situation.

Free-body diagram (FBD): in which only the objects being analyzed is drawn, with arrows showing all the forces acting on the object.

Problems 1. Two men are pushing a stalled car. The mass of the car is 1850 kg. One person

applies a force of 275 N to the car, while the other applies a force of 395 N. Both forces act in the same direction. A third force of 560 N also acts on the car but in opposite direction. This force is due to friction and the extent to which the pavement is opposing the motion of the tires. Find the acceleration of the car. [ ex 1/84/P]

Solution: ∑F = + 275 N + 395 N – 560 N = + 110 a = ∑F / m = + 110 N/ 1850 kg = + 0.059 m/s 2. 2. A man is stranded on a raft (mass of man and raft = 1300 kg). By paddling, he

causes an average force P of 17 N to be applied to the raft in a direction due east. The wind also exerts a force A on the raft. This force has a magnitude of 15 N and points 670 north of east. Ignoring any resistance from the water, find the x and y components of the raft’s acceleration. Reasoning: Since the mass of the raft and man is known, Newton’s second law can be used to determine the acceleration components from the given forces. The acceleration component in a given direction is the component of the net force in that direction divided by the mass

Force

x Component

P

+ 17 N

0

A

(15 N) cos 670 ≈ 6 N

(15 N) sin 670 ≈ 14 N

∑F

∑Fx = 23 N

∑Fy = 14 N

y

Component

ax = (∑Fx)/m = (23 N)/1300 kg = +0.018 m/s2 ay = ∑Fb /m = (14N)/1300 kg = +0.011 m/s2 3. A cannon gives a 5.0-kg shell a forward acceleration of 5.0 x 103 m/s2.Find the magnitude and direction of the net force. ** [2a/73P11]

V. Newton’s Third Law of Motion.

When you push your hand on the desk in one direction, you feel the force of the desk pushing back in your hand in the opposite direction. This means there isn’t just one force; there is a pair of forces. Newton was the first to realize that all forces occur in pairs and there is no such thing as an isolated force, existing all by itself. His third law of motion deals with this fundamental characteristic of forces. Whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body. This law is also known as the “action-reaction” law and is quoted as: “For every action (force), there is an equal, but opposite reaction.’ To avoid confusion, though, remember that the “action” force and the “reaction“ force are acting on different objects. Accelerations Produced by Action and Reaction Forces Example. 1. A 92-kg astronaut drifting just outside an 11,000-kg spacecraft pushes on the spacecraft with a force of +36 N. Find the accelerations of the spacecraft and the astronaut. Reasoning: According to Newton’s third law, when the astronaut applies the force +36 N to the spacecraft, the spacecraft applies a reaction force of -36 N to the astronaut. As a result, the spacecraft and the astronaut accelerates in opposite directions. Although the action and the reaction forces have the same magnitude, they do not create accelerations of the same magnitude, because the spacecraft and the astronaut have different masses. According to Newton’s

second law, the astronaut, having a much smaller mass, will experience a much larger acceleration. On applying the second law, we note that the net force acting on the spacecraft is ΣF = +36 N, while the force acting on the astronaut is ΣF = -36 N. Solution: Using the second law, aS = +36 N /11,000 kg = 0.0033 m/s2 Answer aA = -36 N /92 kg = 0.39 m/s2 Answer Insight: Even though the magnitudes of the action and reaction forces are always equal, these forces do not necessarily produce accelerations of equal magnitudes, since each force acts on an object which may have a different mass. VI. Newton’s Law of Gravitational Attraction Every particle in the universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. This force acts along the line joining the two particles. The magnitude of the gravitational force can be written as

F = G x m1m2_ r2 where m1 and m2 are the masses of the two particles, r is the distance between them and G is a universal constant the value of which was calculated by Newton himself. Verification of this value, however, did not occur until after more than a century later in 1798, when Henry Cavendish, an English physicist, confirmed Newton’s hypothesis and calculated the value of G as

G = 6.67 x 10-11 N-m2/kg2 Examples. 1. A 50-kg person and a 75-kg person are sitting on a bench so that their centers are about 50 cm apart. Estimate the magnitude of the gravitational force each exerts on the other. Solution:

F = (6.67 x 10-11 N-m2/kg2)(50 kg)(75 kg)

(0.50 m)2 = 1.0 x 10-6 N 2. What is the force of gravity acting on a 2000-kg spacecraft when it orbits two

Earth radii from the Earth’s center (i. e., a distance rE = 6380 km above the Earth’s surface)? The mass of the Earth is ME = 5.98 x 1024 kg. Solution: We could plug all the numbers into the equation but there is a simpler approach. The spacecraft is twice as far from the Earth’s center as when at the surface of the Earth. Therefore, since the force of gravity decreases as the square of the distance, the force of gravity on it will be only one fourth of its weight at the Earth’s surface. FG = ¼ mg = ¼ (2000 kg)(9.8 m/s2) = 4900 N = 7.35 x 1022 kg) due to the gravitational attraction of both the Earth (mE = 5.98 x 1024 kg) and the Sun (mS = 1.99 x 1030 kg), assuming they are at right angles to each other.

3. Find the net force on the Moon (m

Solution:

M

First, calculate the magnitudes of the two forces and then add them vectorially.

Distance from Earth to Moon = 3.84 x 108 m Distance from Moon to the Sun = 1.5 x 108 km The force on the Moon due to the Earth is: FME = (6.67 x 10-11 N-m2/kg2)(7.35 x 1022 kg)(5.98 x 1024 kg)_ (3.84 x 108 m)2 = 1.99 x 1020 N The force on the Moon due to the Sun is FMS = (6.67 X 10-11 N-m2/kg2)(7.35 x 1022 kg)(1.99 x 1030 kg)_ (1,50 x 1011)2 20 = 4.34 x 10 N Since the forces act at right angles, the total force is F = √ (1.99)2 + (4.34)2 x 1020 = 4.77 x 1020 N which acts at an angle Ө = tan-1 (1.99/4.34) = 24.60

Distinction between: Newton’s Universal Law of Gravitation: describes a particular force, gravity and how its strength varies with the distance and masses involved. Newton’s Second Law of Motion: relates the net force on a body (the vector sum whatever their sources) to the mass and acceleration of that body. VII. Momentum Momentum:

the strength (or quantity) of an object’s motion. Momentum depends both on the mass and velocity of an object. The faster you throw the ball, the more momentum it has so the more impact it has when it hits something. A bullet has a small mass, but it has great momentum because of its high velocity. A tran, even moving slowly, has a great momentum because of its large mass. p = mv (kg-m/s)

from F = ma, a = (v – v0)/t = m (v – v0)/t = (mv – mv0)/t mv = final momentum mv0 = initial momentum therefore, Impulse:

force is the rate of change of momentum. the interaction that changes an object’s momentum – a force acting for a time interval.Impulse is equal to change in momentum.

Law of Conservation of Momentum

between

Given a system of objects, whenever there is no net force acting on the system from the outside, the force that are involved act only the object’s within the system. The total momentum of the system remains constant.

Example 1. A 150-grain bullet for a 30-06 rifle has a mass m of 0.01 kg and a muzzle velocity

of 900 m/s. Calculate the bullet’s momentum. If the mass M of the rifle is 4.5 kg, what is its speed of recoil? Solution: p = mv = 0.01 kg x 900 m/s = 9 kg-m/s V = p/M = (9 kg-m/s)/4.5 kg = 2 m/s If you do not hold the rifle snugly against your shoulder, the rifle will hit your shoulder at this speed (4.5 mph!) and hurt you. 2. Two cars of mass 1000 kg each collided along a straight line. If the velocity of the two cars locked together was 10 m/s after the crash, find the total momentum after the crash. If only one car was moving before the crash, what was its speed? Soln 3 A freight train is being assembled in a switching yard. Car 1 has a mass of m1 = 65,000 kg and moves at a velocity of v01 = 0.80 m/s Car 2, with a mass of m2 = 92,000 kg and a velocity of v02 = 1.3 m/s, overtakes Car 1 and.couples to it. Neglecting friction, find the common velocity vf of the cars after they become coupled. Reasoning: The two boxcars constitute the system. The sum of the external forces acting on the system is zero, because the weight of each car is balanced by a corresponding normal force, and friction is being neglected. Thus, the system is isolated, and the principle of conservation of linear momentum applies,.The coupling forces that each car exerts on the other are internal forces and do not affect the applicability of this principle. Soln

Forces in a Nutshell (A Review)

Newton’s Laws of Motion

Newton’s Law of Universal Gravitation. Two particles having masses m1 and m2 separated by a distance r, exert on each other a force given by: F = G (m1m2/r2) directed along the line joining the particles. Definitions: inertia:

an object’s resistance to a change in its velocity. Newton’s first law of motion is also known as the law of inertia.

inverse proportionality:

a relationship in which a quantity is related to the reciprocal of a second quantity.

mass:

a measure of the inertia of a body.

weight:

the force of gravity acting on a body.

normal force:

FN, is one component of the force that a surface exerts on an object with which it is in contact – namely, the component that is perpendicular to the surface.

momentum:

the linear momentum of an object is the product of its inertial mass and its velocity.

impulse:

the interaction that changes an object’s

momentum:

– a force acting for a time interval.

VIII. Work and Energy Work done on an object by a constant force (constant in both magnitude and direction) is defined as the product of the magnitude of the displacement and the component of the force parallel to the displacement: W = F’ x d

where F’ is the component of the force F parallel to the displacement d. We can also write: W = Fd cos Ө, where F is the magnitude of the constant force, d is the magnitude of the displacement of the object, and Ө is the angle between the directions of the force and the displacement. Work is a scalar quantity– it has only magnitude. In case the motion and the force are on the same direction ( Ө = 0 and cos Ө = 1), then, W = Fd. For example, when you push a loaded grocery cart a distance of 50 m by exerting a horizontal force of 30 N on the cart, you do 30 N x 50 m = 1500 N-m of work on the cart. In SI units, work is measured in newton-meters, called joule (J): 1 J = 1 N-m. In cgs system, the unit of work is called the erg and is defined as 1 erg = 1 dyne-cm. In British units, work is measured in foot-pounds. It is easy to show that 1 J = 107 erg = 0.7376 ft-lb. When dealing with work, as with force, it is necessary to specify whether you are talking about work done by a specific object or done on a specific object. It is also important to specify whether the work done is due to one particular force (and which one), or work done by the total net force on the object. Example A 50-kg crate is pulled off 40 m along a horizontal floor by a constant force exerted by a person, FP = 100 N, which act at a 370 angle. The floor is rough and exerts a friction force Ffr = 50 N. Determine the work done by each force on the crate, and the net work done on the crate.

Solution: There are four forces acting on the crate: the force exerted by the person FP , the friction force Ffr ; the crate’s weight mg ; and the normal force FN, exerted upward by the floor. The work done by the gravitational and normal forces is zero, since they are perpendicular to the displacement x (Ө = 900): Wg = mgx cos 900 = 0 WN = FNx cos 900 = 0 WP = FPx cos Ө = (100)(40 m) cos 370 = 3200 J Wfr = Ffrx cos 1800 = (50 N) (40 m) (-1) = - 2000 Jt Wnet = 0 + 0 + 3200 J – 2000 J = 1200 J Ans. The net work can also be calculated by first determining the net force on the object and then taking its component along the displacement. (Fnet)x = FP cos Ө – Ffr.

Wnet = (Fnet)x = (FPcos Ө – Ffr)x = (100 N cos 370 – 50 N) (40 m) = 1200 J Ans. Problem

Determine the work a hiker must do on a 15.0 kg backpack to carry it up a hill of height h = 10.0 m. Determine also (b) the work done by gravity on the backpack, and (c) the net work done on the backpack. Assume the motion is smooth and at constant velocity (i.e, there is negligible acceleration).

Solution: (a) the forces on the backpack are mg, the force of gravity acting downward, and, FH , the force the hiker must exert upward to support the pack. Since we assume there is negligible acceleration, horizontal forces are negligible. In the vertical (y) direction, we choose up as positive. Newton’s 2nd law applied to the backpack gives…

Kinetic Energy

A moving object can do work on another object it strikes. A flying cannonball does work on a brick wall it knocks down; a moving hammer does work on a nail it strikes. In either case, a moving object exerts a force on a second object and moves it through a distance. An object in motion has the ability to do work and thus can be said to have energy. The energy of motion is called kinetic energy. Consider an object of mass m moving in a straight line with an initial speed v1. To accelerate it uniformly to a speed v2, a constant net force Fnet is exerted on it parallel to its motion over the distance d. Then the net work done on the object is Wnet = Fnetd. We apply Newton’s 2nd law, which we now write as v22 = v21 + 2ad. We derive, Wnet = ½ mv22 – ½ mv21

[A] Kinetic Energy

We define the quantity ½ mv2 to be the translational (as distinguished from rotational) kinetic energy (KE) of an object. KE = ½ mv2

[B]

Rewriting [A], Wnet = KE2 – KE1 Wnet = ΔKE

or [C]

The net work done on an object is equal to the change in kinetic energy. This is the work-energy principle.

Work:

the energy transferred to an object by an applied force over a measured distance.

Negative Work:

negative work is done if the force is opposite the direction of the displacement. This means that the force tends to decrease the speed of the object.

.

The Work-Energy Theorem

Newton’s second law(F=ma) , the definition of work(W=Fd), and the equations of kinematics(KE=mv*v/2) are brought together to produce the work-energy theorem. Example 1.

KE and work done on a baseball. A 145-g baseball is thrown with a speed of 25 m/s. (a) What is the kinetic energy? (b) How much work was done on the ball to make it reach this speed, if it started from rest?

Solution:

(a) The kinetic energy is

KE = ½ mv2 = ½ (0.145 kg)(25 m/s)2 = 45 J (b) Wnet = ½ mv22 – ½ mv21 2.

Work on a car to increase its KE. How much work is required to accelerate a 1000-kg car from 20 m/s to 30 m/s?

Solution:

The net work needed is equal to the increase in kinetic energy. W = ke2 – ke1 = ½ m2v22 – ½ m1v21 = ½ (1000 kg)(30 m/s)2 – ½ (1000 kg)(20 m/s)2

2 The moon revolves around the Earth in a circular orbit, held there by the Earth’s gravitational force. Does Earth do work on the moon?

Form of Energy

Forms of Energy: Description

Exam

Thermal

The atoms and molecules of a substance possess thermal energy. The faster the atoms move, the greater the thermal energy.

A

Electrical

Possessed by charged particles. The charges are transferred as they move through an electric circuit.

A

Radiant Nuclear potential

Travels by means of waves w/o requiring particles. The nucleus of atoms has stored energy. Released by reactions such as nuclear fission or nuclear fusion.

Gravitational A raised object has stored energy due to its position potential above some reference level. Kinetic Elastic potential

B B C

Every moving object has energy of motion.

C

Energy stored in stretched or compressed objects

C

Sound Chemical potential

Produced by vibrations, travels by waves through a material to a receiver.

D

In chemical reactions, new molecules are formed and energy is released or absorbed.

D

Examples of Forms of Energy: A.

Electrical energy delivered to the stove and heats the water in the pot. Thermal energy in the boiling water transfers to the food to cook it.

B.

The Sun emits radiant energies, such as infrared radiation, visible light, and ultra-violet radiation. The Sun’s energy comes from nuclear fusion reactions n its core.

C. At the highest point above the trampoline, the athlete has the greatest amount of gravitational potential energy. The energy changes to kinetic energy as her downward velocity increases. The kinetic energy then changes into elastic potential energy in the trampoline to help her bounce back up. Chemical potential energy is released when fireworks explode. Some that energy is transformed into sound energy . Energy Transformations: the ability of the forms of energy to from one to another is called energy transformation. E.g., in a microwave oven, electrical energy transforms to radiant energy (microwave), then thermal energy in the food being cooked. C.

Energy Transformation Equation:

Describes an energy transformation using equations with arrows. For the microwave oven example, the equation is:

Potential Energy The energy possessed by an object because of its position. The energy that is stored and held in readiness has the potential for doing work. Examples:

a stretched or compressed spring; a drawn bow, where the energy stored in the bow can do work on the arrow; a stretched rubber band in a sling shot; Chemical energy stored in fuels is also potential energy (actually energy of position at the submicroscopic level.) This energy becomes available when the positions of the

electric charges within and between the molecules are altered, i.e., when a chemical reaction takes place. Potential energy is found also in fossil fuels, electric batteries and the food we eat. Gravitational Potential Energy:

The potential energy due to the elevated position of an object (e.g., water in an elevated reservoir has potential energy).

gravitational potential energy = weight x height

PE = mgh Gravitational Potential Energy The potential energy of an object relative to a reference level is:

Eg = mgh where g = 9.8 N/kg In SI, energy is measured in joules, mass in kilograms, and height (or displacement) in meters. Problem: In the sport of pole vaulting, the jumper’s center of mass must clear the pole. Assume that a 59-kg jumper must raise the center from 1.1 m off the ground to 4.6 m off the ground. What is the jumper’s gravitational potential energy at the top of the bar relative to where he started to jump? Solution: The height of the jumper’s center of mass above the reference level indicated is h = 4.6 m – 1.1 m = 3.5 m m = 59 kg; g = 9.8 N/kg; Eg = ? Eg = mgh = (59 kg)(9.8 N/kg)(3.5 m) = 2.0 x 103 joules Ans. Understanding Concepts

133p11

1. A 485-g book is resting on a desk 62 cm high. Calculate the book’s gravitational potential energy relative to (a) the desktop and (b) the floor. 2. Rearrange the equation Eg = mgh to obtain an equation for (a) m, (b) g, and (c)

h. 3. The elevation at the base of a ski hill is 350 m above sea level. A ski lift raises a

skier (m = 72 kg including equipment) to the top of the hill. If the skier’s gravitational potential energy to the base of the hill is now 9.2 x 105 J, what is the elevation at the top of the hill?{H+350=(Eg/m*g)+350=(9.2 x 105 J/72*9.8)+350} 4. The spiral shaft in a grain auger raises grain from a farmer’s truck into a storage bin. Assume that the auger does 6.2 x 105 J of work on a certain amount of grain to raise 4.2 m from the truck to the top of the bin. What is the total mass of the grain moved? Ignore friction.(Eg/g*h=6.2 x 105 /9.8*4.2) 5. A fully-dressed astronaut, weighing 1.2 x 103 N on Earth, is about to jump down

from a space capsule that has just landed safely on planet X. The drop to the surface of X is 2.8 m, and the astronaut’s gravitational potential energy relative to the surface is 1.1 x 103 J. What is the magnitude of the gravitational field strength (a) on Planet X?{m=G/g=1.2 x 103 /9.8 a=Eg/m*h} How long does the jump take? {t*t=2h/a} What is the astronaut’s maximum speed? (v*v=2ah) 6. How much work is done on a 100-N boulder that you carry horizontally across a

10-m room? How much PE does the boulder gain? 0That you lift 1 m? 100 [107/cp]

Power Power is the rate of doing work or transforming energy

Power, like work, energy and time, is a scalar quantity. In SI unit, the measure of power is watt (W, Joules per second): of power that would raise a1-kg mass (with a weight of 10 newtons) a height of 0.1 meter each second.

horsepower (hp)

550 foot-pounds per second. 746 watts.

Problem: 1. What is the power of a cyclist who transforms 2.7 x 104 J of energy in 3.0 min?

ΔE = 2.7 x 104 J; Δt = 30 min = 1.8 x 102 s;

Given:

P=?

Solution: P = ΔE/ Δt = 2 x 104/1.8 x 102 = 1.5 x 102 W 2. 2. A 51-kg student climbs up a ladder h = 3.0 m in 4.7 s. Calculate the student’s

(a) gravitational climb. Given:

potential energy at the top of the climb, and (b) power for the m = 51 kg; h = 3.0 m; g = 9.8 N/kg; 4.7 s; Eg = ?;

P=?

Solution: (a) (B)

Eg = mgh = 5.1 kg (9.8 N/kg)(3 m) = 1.5 x 103 j P = ΔE/ Δt = 1.5 x 103 J/4.7 s = 3.2 x 102 W

3. 3. A fully out-fitted mountain climber, complete with camping equipment, has a

mass of 85 kg. If the climber climbs from an elevation of 2900 m to 3640 m in exactly 1 hr, what is the climber’s average power? 4. 4. The power rating of the world’s largest wind generator is 3.0 MW. How long

would it take such a generator to produce 1.0 x 1012 J, the amount of energy needed to launch a rocket?Δt = ΔE/P 5. An elevator motor provides 32 kW of power while it lifts the elevator 24 m at a

constant speed. If the elevator’s mass is 2200 kg including passengers, how long does the motion take? Δt = ΔE/P 6. The nuclear generating station located at Pickering, Ontario, one of the largest in

the world, is rated at 2160 MW of electrical power output. How much electrical energy, in megajoules, can this station produce in one day? 7. How much energy is required is to leave a 75-watt yard light on for 8 hours?

8. How much electrical energy does a motor running at 1000 watts for 8 hours require? [149wv]

9. What average power does a weightlifter need to lift 300 lbs a distance of 4 feet in

1.2 s?

[21/155wv]

Consumption of Energy Notes: •

It takes 9.8 J of energy (in the form of work) to raise a 1-kg object a vertical distance of 1 m.

•

Your body may need to consume 4 J, 5 J, or even 10 J to transform a joule of energy.

•

It takes about 200 J to raise your body from a sitting to standing position.

•

You need about 2 kJ to climb a flight stairs.

How much energy, in joules, do you think you consume daily? Your average daily food intake provides about 10 MJ of energy for your activities. In a developed country, the average person consumes about 1GJ (1000 MJ) of energy daily. This is about 100 times (!) the amount we need to survive (10 MJ).

There is a large amount of other energy you use and this includes energy: •

• • • • •

you use to cook your food that lights, heats and cools your home and school for transportation used to make your clothes, books, furniture, CD player, and other appliances used to build and light your streets, to remove your sewage and garbage and on and on…

How much energy did peoples of different eras consume? Daily Average Energy Consumption per Person.

Lifestyle

Era

Type of Energy Used

Daily Consumpt (MJ)

Primitive

Pre-Stone Age Energy from food only

10

Seasonally nomadic

Stone Age

Energy from wood fires for cooking and heating

22

Agricultural

Medieval

Energy from domesticated animals, water, wind, coal

100

Industrial

19th Century

Energy mainly from coal to run industries and steam engines

300

Technological

Present

Energy from fossil fuels and nuclear 1000 sources used for electricity, (in developed transportation, industry, agriculture, countries) etc.

Understanding Concepts 1.

Estimate the average yearly consumption of a developed country with a population of 3 x 107.

2. What fuel source replaced wood as the human lifestyle changed from nomadic to agricultural? 3. What fuel replaced the one you named in (2)? 4. What fuel replaced the one you named in (3)? 5. Why might two countries of similar technological levels have very different per capita energy consumption?

To understand the impact of growth rate, suppose that a town’s annual budget for transportation is $ 100 000.00 With the projected increase in salaries, cost of energy and population, an average expense growth of 8% p.a. is expected. Study the effects of this growth rate in the

table that follows…

The Effects of a Growth of 8 % p.a. on budget Year

Budget

0

$100 000.00

1

$108 000.00

($100 000 x 1.081)

2

$ 116 640.00

($100 000 x 1.082)

3

$ 125 971.20

($100 000 x 1.083)

9

$199 900.46

($100 000 x 1.089)

18

$399 601.95

($100 000 x 1.0818)

27

$ 798 806.15

($100 000 x 1.0827)

36

$ 1 596 817.18

($100 000 x 1.0836)

45

$3 192 044.94

($100 000 x 1.0845)

At 8% growth per annum, the original value has doubled in 9 years. After 45(5*9=45) years, a person’s period of work expectancy, our energy use will be 32 (2 的 5 次方=32)times greater! Note that the product of the growth rate (8% per annum) and the time it takes for an amount to double (in this example, 9 years), can be used to estimate the doubling time for particular growth rates.