This document provides information about physics concepts related to work. It defines work using the formula Work = Force x Distance, and explains this definition in examples of Atlas holding up the Earth and a waiter carrying a tray. It then discusses kinetic energy and the work-energy theorem. Examples are provided for work done by gravity over a change in height, gravitational potential energy, conservation of energy, free fall, pendulums, roller coasters, springs, and multiple forces. Power is also defined and examples using kilowatt-hours are given. The document concludes with multiple choice questions testing understanding of these concepts.

1. PHYSICS
PROJECT
WORK
made by- Archana pal
class-9
roll no.7

2. Physicist’s definition of “work ”
(n A s
ot
ca
a
l
ve ar
ct
or
)
dist∥
dist
Work = F
x
dist∥

3. Atlas holds up the
Earth
But he doesn’t move,
dist∥ = 0
Work= F x dist ∥ = 0
He doesn’t do any work!

4. Garcon does work when
he picks up the tray
but not while he
carries it around
the room
dist is not zero,
but dist∥ is 0

5. or
ct on
ve ti
A qua
e
Why this definition?
Newton’s 2 nd law:
A
eq sca
ua lar
tio
n
F=m a
Definition of work
+ a little calculus
Work= change in ½mv2
This scalar quantity is given
a special name: kinetic energy

6. Work = change in
KE
This is called:
the Work-Energy Theorem

7. Work done by gravity
start
dist
dist∥
change in
vertical height
W=mg
Work = F
=
end
- mg
=
x
x
change in
-change in
dist ∥
height
mg h

8. Gravitational Potential Energy
Work grav = -change in mgh
This is called:
“Gravitational Potential
Energy” (or PEgrav)
changegrav = -change in PEgrav
Work in PE grav = -Work grav

9. If gravity is the only force
doing work….
Work-energy theorem:
-change in mgh = change in ½ mv 2
0 = change in mgh + change in ½ mv 2
change in (mgh + ½ mv2) = 0
mgh + ½ mv2 = constant

10. Conservation of energy
mgh + ½ mv2 = constant
Gravitational
Potential energy
Kinetic energy
If gravity is the only force that does work:
PE + KE = constant
Energy is conserved

11. Free fall
height
(reminder)
t = 0s
V0 = 0
t = 1s
80m
75m
V1 = 10m/s
60m
t = 2s
V2 = 20m/s
t = 3s
35m
V3 = 30m/s
t = 4s
V4 = 40m/s
0m

12. m=1kg free falls from 80m
t = 0s
V0 = 0 h0=80m
mgh
800J
½ mv2
sum
0
800J
50J
800J
t = 1s
V1 = 10m/s; h1=75m
750J
t = 2s
V2 = 20m/s; h2=60m
600J
200J
800J
350J
450J
800J
t = 3s
V3 = 30m/s; h3=35m
t = 4s
V4 = 40m/s; h4=0
0
800J
800J

13. pendulum
T
W=mg
Two forces: T and W
T is always
┴
to the motion

14. Pendulum conserves energy
E=mghmax
E=mghmax
hmax
E=1/2 m(vmax)2

15. Roller coaster

16. Work done by a spring
Relaxed
Position
F=0
F
x
I compress
the spring
(I do + work;
spring does
-work)
Work done by spring = - change in ½ kx2

17. Spring Potential Energy
Workspring = -change in ½ kx2
This is the:
“Spring’s Potential
Energy” (or PEspring)
Workspring = -change in PEspring
change in PEspring = -Workspring

18. If spring is the only force doing
work….
Work-energy theorem:
-change in ½ kx2 = change in ½ mv2
0 = change in ½ kx2 + change in ½ mv2
change in ( ½ kx2 + ½ mv2) = 0
½ kx2 + ½ mv2 = constant

19. Conservation of energy
springs & gravity
mgh + ½ kx2 + ½ mv2 = constant
Gravitational
spring
potential energy potential energy
Kinetic energy
If elastic force & gravity are the only force doing work:
PEgrav + PEspring + KE = constant
Energy is conserved

20. Two types of forces:
“Conservative”
forces
forces that do + & – work
“Dissipative”
•Gravity
•Friction
•Elastic (springs, etc)
•Viscosity
•Electrical forces
•….
•…
-work 
change in PE
forces
forces that only do – work
-work  heat
(no potential energy.)

21. (-)Work done by friction heat

22. Work-energy theorem
(all forces)
Workfric =
Work done
dissipative
Forces
(always -)
change in
(PE+KE)
potential energy
From all
Conservative forces
Kinetic
energy
-Workfric -change in heat energy
Workfric = = change in heat energy
-change
=
(PE+KE)
in Heat Energy
change in

23. Work – Energy Theorem
(all forces)
0 =
change in
0 =
+
(PE+KE)
change in Heat Energy
change in
Heat Energy
(Heat
Energy+PE+KE)
+ PE + KE = constant
Law of Conservation of Energy

24. Energy conversion while skiing
Potential energy
Potential energykinetic energy
Friction: energy gets
converted to heat

25. Power
Rate of using energy:
Units:
Joule
1 second
amout of energy
Power = elapsed time
= 1 Watt
A 100 W light bulb
consumes 100 J of
electrical energy each
second to produce light

26. Other units
Over a full day, a work-horse can
have an average work output of
more than 750 J
oules each second
1 Horsepower = 750 Watts

27. Kilowatt hours
energy
Power = time
 energy = power
 power unit
Elec companies use:
x
Kilowatts
(103 W)
x
time
time unit = energy unit
x
hours
(3600 s)
1 kilowatt-hour = 1kW-hr
= 103 W x 3.6x103 s = 3.6x106 Ws
J
HECO charges us about 15 cents /kW-hr

28. Multiple choice questions
1)The work done by a weight of 1kg mass when it moves up
through 1m is:
(a)10 joule
(b)-10 joule
(c)0.1 joule
(d)-0.1 joule
2)A stone is tied to a string and then whirled in a circle. The
work done on it by the string is:
(a)Positive
(c)zero
(b)negative
(d)undefined
ans -1)-10 joule 2)zero

29. 3)When a force retards the motion of a body, the work done is:
(a)positive
(b)zero
(c)negative
(d)undefined
4)The speed of a particle is doubled. Its kinetic
energy:
(a)remains the same (b)becomes two times
(c)becomes half
(d)becomes four times
ans-3)negative 4)becomes four times

30. 5)A body of mass 5kg falls through a height of 5m. The loss in
potential energy of the mass is:
(a)250 j
(b)25 j
(c)2.5 kj
(d)50 j
6)Potential energy of a person is minimum when:
(a)person is standing
(b)person is sitting in a chair
(c)person is sitting on the ground (d)person is lying on the
ground
ans-5) 250 j 6) person is lying on the ground

31. 7)An electric motor creates a tension of 4500N in hoisting a
cable and reels it at a rate of 2m/s.The power of the motor is:
(a)25kW
(b)9kW
(c)225kW
(d)90kW
8)A engine develops a power of 10kW.How much time will it
take to lift a mass of 200 kg to a height of 40m?(g=10m/s):
(a)4s
(b)5s
(c)8s
(d)10s
ans-
7)9kW
8)8s

32. 9)The power (p) is expressed as:
(a)p=work* distance
(c)p=force* velocity
(b)p=force*distance
(d)p=work/ time
10)1 kWh is equal to:
(a)3.6j
(c)3.6*106 j
10)3.6*106j
(b)3.6kj
(d)36j
ans-9) p=work/time