1© manhattan press (h.k.) ltd. work energy energy 3.6 work, energy and power power power

© Manhattan Press (H.K.) Ltd. 1

• WorkWork

• • EnergyEnergy

3.6 Work, energy and 3.6 Work, energy and powerpower

• • PowerPower


Work

3.6 Work, energy and power (SB p. 125)

Work is done when you lift a load

More work is done

- lift heavier load

- lift load higher

work involves:

- force

- movement caused by force

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More to Know 3More to Know 3


Work


sFW )(Work

W = Fs cos


Work


1. F and s in same direction ( = 0)

W = Fs


Work


2. F and s in perpendicular ( = 90o)

W = 0

Unit – joule (J)1 J = 1 N m

scalar

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More to Know 4More to Know 4


Work


Work done by a variable force

= Area under force-displacement graph

xFdsW 0


Work


e.g.

= Area of triangle under the graph

=

eFdsW 0

Fe21

W = Fe21


Energy


Energy

- ability to do work

Mechanical energy

- energy due to motion, position, physical condition

- kinetic energy or potential energy


Energy


Kinetic energy

The kinetic energy (K or KE) is the energy of a body due to its motion.

W = FsF = ma

v2 = u2 + 2asW = mv2/2

Work done by F is changed to KE of body


Energy


Kinetic energy

Kinetic energy of the body (K) = mv2/2

W = Fs

= (ma)s

= mv2/2 – mu2/2

= Increase in kinetic energy


Energy


Potential energy

An object stores energy due to its physical condition or of its position. The energy stored in the object is called potential energy (U or PE).

E.g.- stretched spring or rubber band- lifting object against earth’s gravity


Energy


Potential energy

Gain in gravitational potential energy= Work done against gravitational force

= Fh

Potential energy (U) = mgh


Energy


Potential energy

Energy stored in stretched spring

= kx2/2k – force constant of spring, x - extension

Obey Hooke’s Law

Elastic potential energy


Energy


Principle of Conservation of Energy

Kinetic energy (K) + Potential energy (U)

= Total energy (E)

K = -U

constant


Energy



E.g.

K = mv2/2K = mv2/2 - 0 and v2 = u2 +2asK = mgh = Loss in gravitational potential energy


Energy



E.g.

Gravitational force acting on man (F) = mgWork done by F (W) = Fh = mgh = K

Work done by gravitational force is converted into gain in kinetic energy


Energy


Relationship between force and potential energy E = K + U K = -U

W = F(s) = -U

Force (F) = dsdU


Energy


Relationship between force and potential energy E.g.

U = mgs

mgmgsdsd

dsdUF )(

F is –ve (it is in –ve direction)Go to

Example 6Example 6

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Example 7Example 7


Power


Power

- rate at which work is done

- rate at which energy is delivered

Average power = interval TimedoneWork


Power


F and s in same direction

Power

tFs

interval Time

doneWork

Power = Fv

F and s at

Power = Fv cos

Unit – watt (W)

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Example 8Example 8

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Example 9Example 9


Power


Efficiency

- percentage of useful work done compared to the input energy

%100machine by the usedEnergy

machine by the done work UsefulEfficiency

Efficiency < 100%



3.1 Newton’s Laws of Motion3.1 Newton’s Laws of Motion

1. Newton’s First Law of Motion states that an object will remain at rest or move along a straight line with constant speed, unless it is acted upon by a force.

2. Momentum = Mass ×Velocity

3. Newton’s Second Law of Motion states that the rate of change of momentum of an object is directly proportional to the net force acted on it, and the motion occurs along the direction of the force.



3.1 Newton’s Laws of Motion3.1 Newton’s Laws of Motion

4. Newton’s Third Law of Motion states that when two bodies interact, they exert equal but opposite forces on each other. They are called action and reaction.

5. Action and reaction are equal in magnitude but opposite in direction. They act on different objects.



3.2 Principle of Conservation of Linear 3.2 Principle of Conservation of Linear MomentumMomentum6. The total linear momentum of a system is

constant, if there is no external forces acting on the system.

7. The Principle of Conservation of Linear Momentum and Newton’s Third Law of Motion are consistent with each other.



3.3 Elastic and inelastic collisions3.3 Elastic and inelastic collisions

8. The most fundamental law in Physics is the Principle of Conservation of Energy, which states that the total energy of a closed system is conserved.

9. For elastic collisions, the total kinetic energy after collision is equal to that before collision. Therefore, kinetic energy, linear momentum and total energy are conserved.



3.3 Elastic and inelastic collisions3.3 Elastic and inelastic collisions

10. For inelastic collisions without external force, kinetic energy is not conserved, but total energy and linear momentum are conserved.



3.4 Inertial mass and gravitational mass3.4 Inertial mass and gravitational mass

11. Inertial mass of a body is a measure of the reluctance to change its state.

12. Gravitational mass of a body is a measure of the gravitational attractive force acted on it (weight of the body).



3.5 Conservation of momentum in two-3.5 Conservation of momentum in two-dimensiondimension13. If two equal masses collide obliquely, the

angle between the two masses after collision would be 90o.



3.6 Work, energy and power3.6 Work, energy and power

14. Work (W) = Fs cosθ

where F and s are the magnitude of F and s respectively and θ is the angle between F and s.

15. Work is done by a force F to stretch a spring to produce an extension of e.

FeW21




16. The kinetic energy (K or KE) is the energy of a body due to its motion.

17. An object stores energy due to its physical condition or because of its position. The energy stored in the object is called potential energy (U or PE).




%100machine by the usedEnergy

machine by the done work Useful

machine a of Efficiency 19.

VelocityForceinterval Time

doneWork power Average 18.


End


A force does no work if:

1. There is no displacement. e.g. No work is done by static friction.

2. The displacement is perpendicular to the applied force. e.g. No work is done by gravitational force if one walks on a horizontal road.

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The sign of work

The sign of work depends on the directions of force and displacement.

1. If their directions are the same, then positive work is done.

2. If their directions are opposite, then negative work is done.

Take an example of lifting a box upwards. The lifting force does positive work since it is acted in the same direction as the displacement. However, the weight of the box does negative work since its direction is opposite to that of displacement.

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Q:Q: A sphere of mass m at the end of a string of length is released when the string is horizontal. The sphere then collides with a block of mass 4m at rests on a smooth table. If the collision is perfectly elastic, find in terms of m, and g,(a) the velocity of the sphere, and(b) the velocity of the block after collision.

Solution



Solution:Solution:The velocity u of the sphere just before collision is given by:

If v1 = velocity of sphere and v2 = velocity of block after collision, then using the Principle of Conservation of Momentum,

mu = mv1 + (4m)v2

u = v1 + 4v2 ..........................(1)By conservation of kinetic energy:

u2 = v12 + 4v2

2 .....................(2)Substituting (1) into (2),

(v1 + 4v2)2 = v12 + 4v2

2

v12 + 8v1v2 + 16v2

2 = v12 + 4v2

2

4v2 (3v2 + 2v1) = 0


gu,mgmu 2 21 2

22

21

2 )4(21

21

21 vmmvmu

(3)32

12 ..................vv

Substituting (3) into (1):

g

gvv

guv

vvvu

252

)253(

32

32

253

53

35)

32(4

12

1

111

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TextText


Q:Q: (a) A ball bearing is released from a height of 1 m from a metal plate. Describe how you would determine the fractional decrease in the kinetic energy of the ball bearing after it hits the plate. What happens to the difference in kinetic energy?(b) A neutron of mass m collides head-on with a carbon nucleus of mass 12m which was initially at rest. If the collision is elastic, find the ratio of the kinetic energy of the neutron after collision to its kinetic energy before collision. Solution



Solution:Solution:(a) Suppose that the ball bearing is released from a height ho, and vo is the velocity of the ball bearing just before it hits the metal plate.Using the Principle of Conservation of Energy,

Let v1 = velocity of ball bearing immediately after collision, h1 = height reached by it after collision.Using the Principle of Conservation of Energy,

Therefore, the fractional decrease in kinetic energy:

That is, the fractional decrease in kinetic energy is determined by measuring the heights ho and h1 and substituting them into equation (3).The loss in kinetic energy is converted into heat and sound during collision.


(1)21

o2

o ..................mghmv

(2)21

12

1 ..................mghmv

(3)

21

21

21

o

1o

o

1o2o

21

2o

..................h

hhmgh

mghmgh

mv

mvmv


Solution (cont’d):Solution (cont’d):(b)

Using the Principle of Conservation of Linear Momentum,mu1 + 12m (0) = mv1 + 12mv2

u1 = v1 + 12v2 ......................................... (1)Since the collision is elastic, kinetic energy of the system is conserved.

substituting (1) into (2), (v1 + 12v2)2 = v12 + 12v2

2

v2(132v2 + 24v1) = 0


(2)12 )12(21

21

21 2

221

21

22

21

21 ..................vvu,vmmvmu

(3)112

12 ..................vv


Solution (cont’d):Solution (cont’d):

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720169121

21

1311

21

collision beforeneutron ofenergy Kineticcollisionafter neutron ofenergy Kinetic

1311

)112)(12( :(1) into (3) ngSubstituti

21

2

1

11

111

.

mu

um

uv

vvu


Q:Q: A toy car which is powered by a motor travels with constant velocity v. The figure shows the variation of the energy supplied by the motor with time t. If the constant force generated by the motor is F, find the value of v in terms of a, b, and F. (Neglect resistance to motion due to friction.)

Solution



Solution:Solution:

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TextText


aFb

Fab

v

Fvab

Fv

Fv

Fv

)(Velocity

graph ofGradient

interval TimedoneWork

work doing of Rate Power


Q:Q: A manufacturer claims that the maximum power delivered by the engine of a car of mass 1 200 kg is 90 kW.(a) Find the minimum time in which the car could accelerate from rest to 30 m s−1.(b) An independent test quotes that a time of 12 s is needed to accelerate the car from rest to 30 m s–1. Suggest a reason for the difference from the time you have calculated.

Solution



Solution:Solution:

(b) There is a difference in time because work is done to increase the kinetic energy of the car, and also work against friction.

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s 610902

)30(1200Power

121 )( taken Time

2interval TimedoneWork Power

21 car by theenergy Kinetic car theof engine by the done Work (a)

3

22

2

2

mvt

tmv

mv

1© manhattan press (h.k.) ltd. work energy energy 3.6 work, energy and power power power

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