7.-OL-Work-Energy-And-Power.pdf

Work ii.
É
→ Work is done when
object moves
=
parallel to The direction of friction = 20N
applied force.
Workdone
against friction
Definition : It is the
product of force W= Fxs
and
displacement parallel to
force.
W -
-
20×2
W =
40 J
W =
Fxs •
SI Unit : Joules ( J)
•
Scalar
Quantity
"
Work done
by
"
force in the direction
F- 50N
of movement
I,
>
"
workdone
against
"
force opposite to
work done by applied force the direction
of
W : Fxs movement .
= 50×2
W =
100J
CLASSMATE
YOUR PARTNER FOR PURSUING EXCELLENCE
Physicsby Kashan Rashid

iii.
friction -
-
low
"
'
I
.
.
.
'
'
'
g-
S
"
-
.
-
"
iz ,m
. E£••• ,
,
-
<
5cm -
i
-
i -
-
i
12cm -
V
weight
-
-40N
'
'
r
, ,
i
'
>
Weight when
something is
moving in a circle
a) Workdone
by gravity about a
point leg.
satellite
orbiting
around Earth ) ,
the centripetal force
W =
Fxs is
acting towards center but the
= 40×0.05
displacement / motion ) of satellite is
W = 2J
perependicutar 190° ) to the
force.
b) Workdone
against friction As
force and displacement are not
parallel to one another
,
NO WORK IS
W =
Fxs BEING DONE
by centripetal force.
= 10×0.13
w =
1.3J
CLASSMATE

Energy sources of Energy
The
ability of a
body to do 1 .
Non-
Renewable Resources
work.
→
Consumption rate
of these resources
are
extremely high compared to
•
SI Unit:
Joules 1J ) °
Scalar
quantity
the
regeneration rate .
Types Forms
of energy
Coal Oil Natural Nuclear
Gas
1. Mechanical Energy
a) Kinetic Energy 2 . Renewable Resources
b) Potential Energy → Resources
regenerate at an
2. Heat Energy incredibly fast rate so consumption
3. Chemical
Energy doesnot
effect Their reserves.
4 . Sound Energy
5.
light Energy o
Solar ° Wind . Wave
6. Nuclear Energy • Tidal °
Geothermal ☐
Biogas
7. Internal Energy
CLASSMATE

Gravitational Potential Energy Example # I
m -
-200g g-
-
9.8141kg 2,0%0=0-2kg
The
energy stored in a
body due
^
to its
position in a
gravitational Ep
-
-
mgh
field .
4µm =
1002119.811401
Ep =
78.4J
•
It is due to the
pull
• - -
v
of gravity on the Wv
µ
object in the
field .
Example # 2
.
mass
-
-
0.5kg g-
-9.81N/kg
a-
- - - -
•
Weight pulls the
ground Calculate the loss in
object downwards ,
so when the
objection -
-
Grau. Potential Energy
f-
- - - - -
i.
is released ,
it moves .
Hence it has
| am
of the ball .
some
potential .
Ep =
mgh
>
h
"Ñt A- Ep =P-
Ef -
P-
Ei
IEp=mgDh
Potential
'
✓
=
might -
mghi DEp= 0.5×9.8×6
energy mass
'
grainfield strength =
mg tht - hit
= 0.5×9.8×(4-10)
B. Ep = 29.45
DEP = -29.4J
CLASSMATE

kinetic
Energy "
÷
"
kinetic > motion m v E
•
2m ZV E- ✗ 2×22 =
BE
•
The
energy possessed by a
body .
mz
Zv E ✗ 22=2 E
due to in motion .
| |-2
I E
EK =
Izmit >
speed
•
MI 3 2×32
=
¥8
E ✗ 3
L
zz
=
3E
Kinetic v 2 4
energy mass
EKXM EKXVZ
mass KE speed K .E
Example # I
Ek=
tgmv
"
m E v E > v
.
8m15
2m 2E 2v 12T¥ 4E -
-12160011812
3m 3E 3v (3)
'
9E m
-
-600kg
my} Ets 42 125=>44 Ek= 19200J
Mla 1=12 V3 (3)
'
Elq
CLASSMATE

Example # 2 mass
-
-
0.6kg Law
of conservation
of Energy
initial calculate the
gain
5m15 -
in K.E. Kenagy can neither be created
,
nor
L
final -
÷
be
destroyed .
9T is
only
converted
8m15
from one
form
to another .
<
vi.
final speed
•
Total
energy of an entire closed
AEK =
Kott - K -
Ei u : initial
speed system is conserved .
L
=
Izmit
-
Izmit
☐ Ek =
tgm /v2 -
W ) v2
.az ( v
. up
closed
system means
no
energy enters
or leaves the zone .
AEK -
-
1210.61/82-54 1 .
Burning of coal
☐ Ek = 11.7J Chemical
>
Heat
+ Light
energy energy energy
+
produced to
gather
> converted into
CLASSMATE

2.
Object falling from a
height 4. Ball
colliding
with the
ground and
( no air resistance) Ep
-
-100J
rebounding Ino loss
upon
collision )
Ek= 0
mV
Gravitational kinetic .
'
'
'
.
Potential >
energy
.
'
'
Ep-
-60J
i
Energy Ek-
-
40J
Kinetic
energy
>
Elastic
"
"
"
"
potential
>
kinetic
Ep -0
energy
VVEK -100J energy
1 ball bounces off with same speed )
3.
Object falling from a
height
1 with air resistance )
Ep
-
-100J rebounding (
energy
lost on collision )
'
'
'
Ep=goj§
" % " """
Y -
with the
gwun
, any
GPE > Kot + heat Ek -
- O
l
i
'
'
'
.
qv ( less
because
of peed )
air resistance Ek=3OJÉ
Kinetic
>
Elastic RE
"
"
"
Ep:O
'
if
•
energy + heat
vv Ex -
- 80J IÉ • Elastic P-
E > K - E + heat
CLASSMATE

6.
Energy conversion in cell & circuit
Example # I u=o
A ball is released
from rest
,
40m
-
inimiiaiéneigy's ,
on above the
ground .
Ignoring
→
[nside
"- - - - -
"
air resistance ,
calculate the
= in cell / battery =
I v
speed of ball as it hits the
-
Electrical
energy
?
4µmground.
"
l l l l l
✓
l l l l l l l
'
,
,
,
i
,
,
1
,
P. E → K . E
light energy + heat
energy
B v
Potential
energy =
Kinetic
energy
at A at B
7. Solar Panels
nxgh =
Izmit
light energy > Electrical energy 190811401 =
Iz
v2
8. Wind Turbine +
Generator V= ✓ 2×9.8×40
✓ =
28m15
Kinetic
energy >
KE
of >
Electrical
in wind blades
energy in
If energy loss would have occurred,
+ heat
generator v would be less than 28
Mfs.
+ neat
CLASSMATE

,
^
Power
•
Rate at which work is done .
m
4 m¥,
h
¥
•
Rate at which
energy
is converted .
running walking
P -
- Work P=
Energy ☐ same
gain
in P-
E in both cases
time time o
less time taken while running .
-
P=W_ P=
Ey
☐ more
power developed while
running,
t than while
walking .
•
SI Unit : Walt IWI °
Scalar
quantity Example # 2
lamp :P = 100W 1100J/s ) A motor
pumps 20kg of water to a
100J
of electrical
energy
is converted
height of 40m above
ground in 5min .
in 1s .
Assuming no
energy
loss,
calculate The
power rating of motor.
A more
powerful machine can do
the same work or can convert RE -
-
mgh Power-_
E- =
7840
same
energy
is less time .
=
20×9.8×40 t 5×60
P. E =
7840J Power =
26.13W
CLASSMATE

Efficiency
Ratio
of useful output to the
required input.
•
efficiency =
useful output
required input
✗ 100
•
input =
output + waste
eft
-
-
ww÷ eff
-
-
E€,
eft-
k
Pi
eff
-
-
WOE
;
ebb =
.
eff
-
- E
Ei
ldifferent units )
CLASSMATE

1.
2.
3.
Work, energy and Power
Past Papers - MCQs
138
0
-
✓
2000N
→
weight W= Fxs
0 =
2000×0.8
= 1600J
eff
-
-1%0×100
input → ←
output
0
CLASSMATE

10
11
12
13
140
0
t
'
✗
¥:
:
n
108-006=1 -2m
?⃝2m W : Fxs
=
50×1-2
=
60J
0
CLASSMATE

54.
55.
56.
149
W = Fxs
=
300×5
W =
1500J
O
O EE -
-
mgh
"
☒
CLASSMATE

© UCLES 2018 5054/12/O/N/18
6
78 The diagram represents a geothermal power station.
cold
water
steam and
hot water
Which useful energy transformation is taking place?
A electrical energy → potential energy
B electrical energy → thermal energy
C potential energy → electrical energy
D thermal energy → electrical energy
79 A motor is used to lift a load 0.50m vertically, as shown.
0.50m
load
40N
motor
The load weighs 40N. The power of the motor is 20W and the system is 25% efficient.
How long does it take to raise the load?
A 0.040s B 0.25s C 4.0s D 40s
155
Electrical
heat
0
F-
It
or F- E-
t
8 3 .
P=
WI
1 .
Work =
f- ✗ S 20 =
Got
= 40×0.5
=
20J - .
1- =
8¥
2.
efficiency =
:# ✗ too
t -
-
us
25 =
21×100
inp
input =
202-050=805 - .
time ?
- i
0
CLASSMATE

Answers
1.D 2.B 3.A 4.C 5.C 6.D 7.A 8.A 9.C 10.C 11.D 12.C
13.B 14.D 15.B 16.D 17.D 18.A 19.D 20.C 21.D 22.C 23.C
24.D 25.B 26.D 27.B 28.B 29.B 30.B 31.D 32.A 33.D 34.B 35.B
36.D 37.A 38.A 39.A 40.B 41.A 42.B 43.D 44.B 45.C 46.A 47.A
48.D 49.C 50.A 51.B 52.C 53.A 54.C 55.B 56.C 57.B 58.D 59.C
60.A 61.B 62.B 63.C 64.C 65. 66. 67. 68. 69. 70. 71.
72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82.
157
CLASSMATE

160
Kinetic
energy of hands convert to heat
energy
( due to
friction)
there is more
friction force when
pressed harder
-
hence more heat
produced .
=
W =
f- ✗ s 0096J
☒
1
= 1.2×0008 2 J -
N
W = 0.096
a = 2-
11 movement )
0.096
n =
20.8
21
=
P=Wµ°e
=0;?Y-
=
0.48W
0.48W
CLASSMATE

6
163
Chemical
energy of
the
cyclist is converted
Gravitational
Potential
Energy.
Kinetic
energy remains same as
speed is constant .
the potential energy gained
at the top of
the
hill is converted to kinetic
energy as it descends
and then to heat as it brakes to
stop.
P . E =
mgh
5400 = 60×10 ✗ h
h = 9m
9m
CLASSMATE

14
174
W -1540N
W= F ✗ s
=
Weight ✗ h
=
(54×10)×2.8
=
1512 J
1512J
p=Wo
time
=
15¥ = 504W
504W
some energy is converted to kinetic
energy of bricks .
Some energy
is lost as heat to
surrounding.
CLASSMATE

17
179
Solar
wind
E-
-
(6.8×109)×11×365 ✗ 24×60×60)
p=
¥ E-
-
2.14×10
"
J
E-
-
pxt
2. 14 ✗ 1017J
The
quantity of
water varies
Throughout The
air due to
rainfall or siltation.
CLASSMATE

180
-
P-
E =
mgh
=
106×1010×10×170
=
2.72×1013 J
2.72×10135
P=¥= 2.72×101-3 off
-
-
Pont =
6.8×10
"
1×60×60 Pin ¥109
✗ 10°
P =
7.56×109 W
= 89.9
9 90%
input power
some energy
is lost as heat
some energy
is converted to K .E
of
water .
CLASSMATE

2
3
4
5
Answers - Theory
195
<
CLASSMATE

6
7
8
196
CLASSMATE

10
197
CLASSMATE

11
12
13
14
15
198
CLASSMATE

16
17
199
CLASSMATE

18
19.
20.
200
CLASSMATE

21.
22.
23.
24
201
CLASSMATE

25
26
202
CLASSMATE

7.-OL-Work-Energy-And-Power.pdf

Recommended

Recommended

More Related Content

Similar to 7.-OL-Work-Energy-And-Power.pdf

Similar to 7.-OL-Work-Energy-And-Power.pdf (20)

Recently uploaded

Recently uploaded (20)

7.-OL-Work-Energy-And-Power.pdf