Unit 5 Work Energy and Power AS/A Level Physics Cambridge CAIE 9702

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everything on the syllabus like you can

see in the figure

work done is the product of force and

displacement moved in the direction of

the force it is a scalar quantity and

its unit is Jewel the formula of the

work done is W equals

FD where W is work done in Jewel f is

the force in Newtons and D is the

displacement moved in the direction of

the force in meters when the force and

the displac M are in the same

directions in this case the work done is

positive if a 5 Newtons Force acts on a

box causing it to move 10 m in the same

direction are shown the work done can be

calculated using W =

FD substituting F = 5 Newtons and D = 10

m we get the work done equals 50 Jewel

when the force and the displacement are

in the opposite

directions in this case the work done is

negative if a 5 Newtons Force acts on a

box to the left while the Box moves 10 m

to the right are shown the work done can

be calculated using wal

FD substituting F = 5 Newtons and D =

-10 M we get the work done equals -50

Jew the negative sign indicates that the

Force opposes the

displacement when the force F acts on a

box at an angle Theta to the horizontal

causing the box to move D meters to the

right are shown the force F can be

resolved into horizontal and vertical

component the horizontal component is f

cos Theta and the vertical component is

f sin Theta F cos Theta is in the same

direction to the displacement D this

component does work W on the box as F

cos Theta * d f sin Theta is

perpendicular to the displacement D this

component does not work on the

box when a person walks 10 m while

holding a box weighing 10 Newtons the

person exerts a force of 10 Newtons to

hold the

Box however the Box moves forward in a

direction perpendicular to this Force

are shown therefore the work done by the

person holding the box is zero when the

gas in the cylinder does work on a

piston of cross-sectional area a the

pressure p on the Piston has a constant

value the force F due to the gas

pressure acting on the Piston causes the

Piston to move outward with distance

Delta X the work done W by the gas is

given by w = f Delta X the force F

exerted by the gas is equal to to the

pressure p multiplied by the area a as

FAL PA a since a Delta X is the change

in the volume of the cylinder Delta V

therefore the work done by the gas can

be expressed as W equals PTA

V exam style question one a team of nine

dogs can pull a sledge with a combined

force of 800 Newtons at a speed of 1.5 5

m/s for 360 minutes what is the average

work done by each dog during this time

the work done can be calculated using W

equal

FD the force F of nine dogs equals 800

Newtons the distance D can be calculated

using D equals speed V multiplied by the

time T substituting V = 1.5 m/s and t

equal 360 * 60

seconds so the distance D equals

32,400

m substituting F = 800 Newtons and D =

32,400 M we get the work done by nine

dogs equals

25,920 th000 Jewel so the work done by

one dog equals

25,920 th000 / by 9 is equal to 2.9 *

10^ 6 Jew for two significant

figures exam style question two a piston

in a gas supply pump has an area of 600

square cm and it moves a distance of 40

cm during one stroke the pump moves the

gas against a fixed pressure of 5,000

pascals how much work is done by the

Piston during one stroke the work done

by the gas can be calculated using W =

FD the force F by the gas can be

calculated using F equal p a the

pressure p by the gas equals 5,000

pascals the area a equal 600 square

cm we convert the area a into square m

like this which is 0.06 s m the distance

D equal 0.4 m M substituting P = 5,000 a

=

0.06 and D =

0.4 we get the work done by the gas W

equals 120

Jew energy is the ability to do work it

is a scalar quantity and its unit is

Jewel there are many form of the energy

as follows kinetic energy is the energy

due to the movement of a mass

gravitational potential energy store in

a mass due to change in its position

within a uniform gravitational field

strength elastic potential energy or

stain energy store in an object due to

change in its shape chemical potential

energy electrical potential energy is

the work done by the electric force due

to the electric field thermal or heat

energy

radiation energy for example light

infrared or gamma Sound Energy nuclear

energy principle of conservation of

energy the energy cannot be created and

destroyed but can only be transferred

from one form to another principle of

work energy work done and the energy

principal state that mechanical or

electrical work done is equal to the

energy transferred

derive the formula of the kinetic energy

as EK equal half of m² v a force F

pushes a box along a smooth surface

causing a displacement of s the Box's

velocity increases from an initial value

of U to a final value of V since the

surface is smooth there is no thermal

energy

loss therefore all the work done W by

the force f is transferred to an

increase in kinetic energy Delta EK K

the work done by the force f w equal

FS so Delta EK k equal FS the force F

equal m a from the equation V ^2 = U ^2

+ 2 a s so a s = half of V ^2 - U

^2 substituting a s this equation like

this so Delta e k = M2 of V ^2 - U ^2

assuming that the initial velocity U

equals 0 therefore Delta EK k equal half

of m² V derive the formula of the

gravitational potential energy as EP

equal mg Delta h a box with a mass m

kilg being lifted from the ground to a

higher level with the height Delta

H since we use the force F that equals

Box's weight to hold the box so f equals

mg therefore all the work done W by the

force f equals mg is transferred to an

increase in gravitational potential

energy Delta

ep the work done by the force f w equal

FS so Delta EP equals FS the force f

equals

mg the distance s equal Delta

H therefore Delta = mg Delta

H exam style question three when a

horizontal Force f is applied to a

frictionless trolley over distance s the

kinetic energy of the trolley changes

from 4 Jew to 8 Jewel if a force of 2f

is applied to the trolley over a

distance of 2s what will the original

kinetic energy of 4 Jewel become the

frictionless surface means that no heat

energy loss or no work done against

friction so the work done by the force f

equals the kinetic energy gain the work

done by the force f = f * s and EK gain

equal 8 J - 4 Jew so FS equals 4 Jew if

a force of 2f is applied to to the

trotle over a distance of 2s the

original kinetic energy of 4 Jewel

become the final kinetic energy

EK so 4 FS = EK - 4 JW and EK = 4 FS + 4

substituting FS equal 4 we get the final

EK equal 20 Jew exam style question 4 a

constant force of 9.0 k

parallel to an inclined plane moves a

body of weight 20 kons through a

distance of 40 m along the plane at

constant speed the body gains 12 m in

height as shown how much of the work

done is dissipated as heat the Box moves

at constant speed to indicate that its

kinetic energy Remains the Same so the

work done by 9 Kon Force equals the

gravitational potential energy gain Plus

work done against friction or dissipated

heat energy therefore f * s = mg Delta h

plus dissipated heat substituting F =

9,000 Newtons mg = 20,000 Newtons Delta

H = 12 M we solve the dissipated heat

equals

120,000

Jew exam style question five a uniform

solid cuboid of concrete of Dimensions

0.5 M * 1.2 M * 0.4 M and weight 4,000

Newtons rests on a flat surface with the

1.2 M Edge vertical as shown in diagram

1 what is the minimum energy required to

roll the cuboid through 90° to the

position shown in diagram 2 with the 0.5

M Edge vertical the solid cuboid will

topple over over if a vertical line

drawn downward from its center of

gravity falls outside its base area are

shown so the minimum energy required to

rise the cuboids center of gravity is mg

Delta H therefore Delta H equals the

difference between the center of gravity

A and B this distance is half of 12 m

equals 06 M the distance X is half of

the diagonal of cuboid this distance is

half of 0.5 m equal 0.25 M the distance

X can be calculated using the

Pythagorean theorem like this we solve

the distance x = 0.65

M so Delta H =

0.65 - 0.6 is equal to

0.05 M substituting mg = 4,000 Newtons

and Delta H equal

0.05 M we get the minimum energy

required equals 200

juwel exam style question six a box of

weight 200 Newtons is pushed so that it

moves at a steady speed along a ramp

through a height of 1.5 M the ramp makes

an angle of 30° with the ground the

friction force on the box is 150 Newtons

while the box is moving what is the work

done by the person the Box moves at

constant speed to indicate that its

kinetic energy Remains the Same so the

work done by the person equals the

gravitational potential energy gain Plus

work done against friction or dissipated

heat energy therefore the work done by

the person equals mg Delta H + friction

time distance D therefore distance D is

the distance along the ramp like this so

sin 30 equal Delta h / D we solve the

distance D = 1.5 / sin 30 = 3

m substituting mg = 200 Newtons Delta H

= 1.5 M friction equal 150 Newtons and D

equal 3 m

we get the work done by the person

equals 750

Jewel exam style question seven a

trolley runs from P to Q along a track

at Q its potential energy is 50 KJ less

than at P at P the kinetic energy of the

trolley is 5

KJ between p and Q the work the trolley

does against friction is 10

kilj what is the kinetic energy of the

trolley at Q the kinetic energy of the

trolley at P equal 5

K when the trolley runs from P to Q the

work done against friction or heat

energy loss is 10 kle and the kinetic

energy at Q is EK K therefore the

gravitational potential energy loss from

P to Q equals 50

k so the gravitational potential energy

loss equals the kinetic energy gain Plus

work done against

friction substituting EP loss equals 50

KJ EK gain equals EK - 5 K and work done

against friction equals 10

K we solve the EK at qal 45

KJ power is the work done per unit time

or energy transferred per unit time time

it is a scalar quantity and its unit is

Watts or Jew per second the formula of

the power P equal W / T or E / T where p

is the power in watts w is work done in

Jewel T is time in seconds and E is the

energy in Jewel if we substitute work

done W equal force f * displacement D

into p = w / t where D / T is speed V so

the power P equals f * V like this exam

style question 8 an area of land is an

average of 2 M below sea level to

prevent flooding pumps are used to lift

rainwater up to sea level what is the

minimum pump output power required to

deal with 1.3 * 10^ 2 kg of rain per day

the power can be calculated using P

equal e / T where the energy E equals

gravitational potential energy mg Delta

H the mass m equal 1.3 * 10^ 9

kg the gravitational field strength g =

9.81 m per

squs the Delta h = 2 m the time T is 1

day converting into second which is

86,400 seconds

substituting m g Delta H and T in the

equation like this we get the power P

equals 300 kilow for two significant

figures exam style question n the

diagram shows a lift system in which the

elevator Mass M1 is partly

counterbalanced by a heavy weight Mass

M2 at what rate does the motor provide

energy to the system when the elevator

is rising at a steady speed v g equals

acceleration of Free Fall the system

moves at steady speed to indicate that

its kinetic energy Remains the Same so

the power input by the motor equals the

rate of gravitational potential of the

system loss when the elevator moves

Upward at steady speed V with distance

Delta H so its kinetic energy is

constant and its gravitational potential

energy gain as M1 G Delta H the mass M2

moves downward at steady speed V with

distance Delta H so its kinetic energy

is constant and its gravitational

potential energy loss as m2g Delta

H so the power input by the motor equals

m2g Delta hus M1 G Delta h / T we factor

out G Delta Delta H like this the Delta

H over T is the steady speed V so the

power input by the motor equals M2 minus

M1

GV exam style question 10 a mass of 2 kg

rests on a frictionless surface it is

attached to a 1 kg mass by a light thin

string which passes over a frictionless

pul

the 1 kg mass is released and it

accelerates

downwards what is the speed of the 2 kg

mass as the 1 kg Mass hits the floor

having fallen a distance of 0.5

M this question can solve using the suat

equations or the principle of

conservation of energy but I will solve

using the principle of conservation of

energy when a 1 kg Mass moves downward

0.5 m causing its speed to increase from

0 to V this causes a 2 kg Mass also

moves 0.5 m to the right and its speed

also increases from 0er to V the

frictionless surface causes no heat loss

or no work done against

friction therefore the gravitational

potential energy of system loss equals

the kinetic energy of system gain the EP

loss equals mg Delta H and EK gain equal

half m² V - half m² U A 1 kg Mass moves

downward causes EP loss equal 1 * 9.81 *

0.5 both masses increase its speed from

0 to V to cause EK gain equal half of 1

+ 2 * Square V - 0 we solve the speed V

equal 1.8 m/s for two significant

figures exam style question 11 an

escalator is 60 M long and lifts

passengers through a vertical height of

30 m as shown to drive the escalator

against the forces of friction when

there are no passengers requires a power

of 2

kilow the escalator is used by

passengers of average mass 60 kg and the

power to overcome friction remains

constant how much power is required to

drive the escalator when it is carrying

20 passengers and is traveling at 0.75

m/ seconds the escalator moves at

constant speed 0.75 m/s to indicate that

its kinetic energy Remains the Same

therefore the power drive the escalator

and passengers equals power when no

passenger plus power drives 20

passengers the the power drives 20

passengers equals work done FD divided

by time t f equals the total weight of

20 passengers equal 60 * 9.81 * 20 is

equal to

11,720

newtons D equals the vertical height of

the escalator equal 30 m t equals the

time of the escalator moving at

0.75 m/s for 60 m

so T = 60 /

0.75 is equal to 80

Seconds substituting f =

11,720 d = 30 and T = 80 we get the

power to drive 20 passengers equals

44145 wats therefore the power drives

the escalator and passengers equals

2,000 +

4,44

14.5 we get the results is 6.4

kilow exam style question 12 a conveyor

belt is driven at velocity V by a motor

sand drops vertically onto the belt at a

rate of M

kg/s what is the additional power needed

to keep the conveyor belt moving at a

steady speed V when the sand starts to

fall on it when sand drops vertically

onto the belt as shown the belt exerts a

force F on the sand and the sand exerts

an equal and opposite Force F on the

Belt the additional power needed to keep

the conveyor belt moving at a steady

speed V to calculate using p equals f *

V the force F can be calculated using

mvus mu U / T the initial velocity U of

the sand equals zero the final velocity

V of the sand equals V the sound's mass

m equal m and the time T = 1 second

substituting all quantities in the

equation like this we get the force FAL

MV substituting FAL MV and V equal V we

get the additional power equal m²

V efficiency is the ratio of the useful

energy or work or power output from the

system to the total energy or work or

power input exam style question

13 a constant force F acting on a car of

mass m moves the car up the slope

through a distance s at constant

velocity V the angle of the slope to the

horizontal is Alpha which expression

gives the efficiency of the process the

efficiency of this question can be

calculated using the efficiency equals

useful energy or work output divided by

total energy or work input the useful

energy output is the gravitational

potential energy gain of the car equals

mg Delta H the Delta H is the vertical

height and sin Alpha equal Delta h / s

so Delta h = s * sin

Alpha therefore EP gain equals MGS sin

Alpha

total energy input equals the work done

by the force F so the work done by the

force f equals f * s therefore the

efficiency equals MGS sin Alpha / FS we

cancel out s like this since the result

is mg sin Alpha over

F exam style question 14

a bow of mass 400 G shoots an arrow of

mass 120 g vertically upwards the

potential energy stored in the bow just

before release is 80 Jewel the system

has an efficiency of

28% what is the height reached by the

Arrow when air resistance is neglected

the efficiency can be calculated using

the efficiency equals useful energy

output divided by total energy input

the efficiency equals

28% the useful energy output equals GP

gain equals mg Delta H the total energy

input equals the potential energy

storing in the bow GP gain equals 0.12

kg * 9.81 * Delta H the potential energy

storing in the bow equals 80 Jewel we

solve the Delta H equal 0 M for two

significant

figures exam style question

15 water from a reservoir is fed to the

turbine of a hydroelectric system at a

rate of 500

kg/s the reservoir is 300 m above the

level of the turbine the electrical

output from the generator driven by the

turbine is 200 amp at a potential

difference of 6,000

volts what is the efficiency of the

system

the efficiency of this question can be

calculated using the efficiency equals

useful power output divided by total

power input the useful power output

equals the electrical power equals VI

the total power input equals the

gravitational potential energy per

second of the water the electrical power

equals 6,000 volts time 200 amp the

gravitational potential energy per

second of the water equal 500 kg/s *

9.81 * 300

M therefore the efficiency equals 82%

for two significant

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