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Form 4© COPYRIGHT BY SUPER EDUCATION GROUP (JO
2.10 – 2.12
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F
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F&'
F
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UNDERSTANDING WORK, ENERGY AND EFFICIENCY
A Work
1. Work is the product of applied force and dis!"#ement in the direction of the applied force.
2. When the work is done ener$y is transferred from one object to another.
3. The work done is equal to the amount of ener$y transferred.
4. The SI unit for work is %o&!e.
The formulae of work
WORK = FORCE X DISPLACEMENT
W = F ! s
No work is done w'en(
)* The object is stationary
2) The direction of motion of the object is perpendicular to that of the applied force
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W : work in Joule/J
F : force in Newton/N
s : displacement in
Work done
Force and displacement in the samedirection
Force and displacement in different directions
W = F.s W = Work F = Force s = displacement
W = FX . sW = F s cos θ W = work F = force s = displacement
θ = angle between forceand displacement
F
F Y F Y
F
F&
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Diagram (a) Diagram (b)
Diagrams (a) and (b) shows a boy pshing a load and a weightlifter lifting aload of !" kg
a) #alclate the work done
i. by the boy
W = F.s = 20 x 2 = 40 Nm or 40
ii. by the weightlifter
W = F.s = m!h
= "0 x #0 x 2 = #200 Nm or #200
$. %&man is plling a bo' with a force of " at an angle of !"o from the hori&ontal.#alclate the work done to mo*e the bo' to a distance of + m.
Wor$ = %omponent of force x displacement&'n the direction of displacement)
= (0 cos "0o x = 2( x = *(
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"!ample 1
"!ample 2
#$ %
&$$
s ' ( m
"!ample 3
"!ample 4
) ' &$$ %
S ' $.# m
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W = Fs
If, F = 40 N and s = 2 m
Hence, W = 40 2
= !0 J
For+$ F
s
W = Fs
= !0 cos "00 #$%
= !0 #0&$% #$%
= 200 J
' '
F = (0 N
) = *&$ m
W = F s = F )
= (0 #*&$%
= 4$&0 J
W = F s
= "00 0&!
= 4!0 J
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%
For+$ F
2.10 – 2.12
Ener$y
1. "ner*+ is the potential to do work.
2. "ner*+ cannot be created nor be destro+ed.
3. "!ist in ,arious forms such as otenti"! ener$y, kineti# ener$y- electrical ener*+- sound ener*+-
nuclear ener*+- heat and chemical ener*+
"ner*+ is de+ined as the capacit+ to do work.
Work is done when ener$y is #onerted from one form to another.
The unit of work is Nm or -o&!e.-*
A/Work done "nd t'e #'"n$e in kineti# ener$y
1. inetic ener*+ is ener*+ of an object due to its motion&
2. /efer to the fi*ure abo,e-
3. "!ample 0 small car of mass 1$$ k* is mo,in* alon* a flat road. The
resultant force on the car is 2$$ %.
a What is its kinetic ener*+ of the car after mo,in* throu*h 1$ m b What is its ,elocit+ after mo,in* throu*h 1$ m
+olution : i-en : m = *00 k. , F = 200 N
a& inetic ener., 1k = Fs
= 200 *0= 2000 J
& 3elocit, - 5 m- 2 = 2000
- = "&(2 m s
6*
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Work ' )s
' mas
' m ,2
The formulae of inetic ener*+- "k ' m,2
Through, v 2 = u2 +2as
u = 0
and, as = ½ v 2
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0/ Work done "nd $r"it"tion"! otenti"! ener$y
h ' 1.( m
1. 5ra,itational potential ener*+ is ener*+ of an object due to its position. possessed b+ an object due to
its position in a *ra,itational field
2. /efer to the fi*ure abo,e
W ' F s ' m$ h
where- F 1 m$
So, Gr"it"tion"! ener$y, E 1 m$'
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2rin#i!e o+ #onser"tion o+ ener$y
the principle of conser,ation of ener*+.
1. "ner*+ cannot be created or destro+ed but can be chan*ed from one form to another form.
2. E3"m!e 6 a thrown ball upwards will achie,e a ma!imum hei*ht before chan*in* its direction and falls
3. "!ample in calculation 6 0 coconut falls from a tree from a hei*ht of 2$ m. What is the ,elocit+ of
coconut just before hittin* the earthi-en : ) = 20 m, u = 0 , . = 7&! ms62 , - = 8
1 p = 1k
m.) = 5 m- 2
m#7&!%#20% = 5m- 2
- 2 = (72, - = *7&! m s6*
2ower
1. 7ower is the rate of doing or!"
rate of energ# transfor$ation"
')erefore, power, 9 =ti$eta!en
or!done
, so, 9 =t
W
W)ere, 9 : power in watt/W
W : work in oule/J
t : time to do work in seconds/s
2. 8nit6 9oule per second 9 s:1 or Watt W
3. 0 wei*htlifter lifts 1#$ k* of wei*hts from the floor to a hei*ht of 2 m abo,e his head in a time of $.# s.
What is the ower $ener"ted b+ the wei*htlifter durin* this time
* ' ;.# ms:2
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;aimum 9otential ener.
inetic ener. decrease potential ener. decrease
and potential ener. and kinetic ener.
Increase increase
;aimum
kinetic ener.
+olution : i-en : m = *!0 k., ) = 2 m, t = 0&! s and . = 7&! ms62& 9 = 8
9 =t
W
=t
$gh
=
0.8
2 9.8 180 ××
= 4 4*0 W
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E++i#ien#y
1.
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2rin#i!e ++ #onser"tion o+ ener$y
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,amy releases a $ kg metal ball from a bilding -" m high (ake accelerationde to gra*ity as 1" ms/$)
a) %t the height of -" m0 the metal ball has (!ra+itational potential ener!ykinetic energy)
b) 2st before the metal ball hits the grond0 the ma'imm energy that it has is (gra*itationalpotential energy$inetic ener!y).
c) #alclatei) he energy of the metal ball at the height of-" m.
,!ra+itational = m!h
= &2) ) &40)
= -00
ii) the kinetic energy of the metal before it hits the grond.
,$inetic = ,!ra+itational= -00
d) What is the principle sed in c ii)3
The principle of conser+ation of ener!y
. % motor lifting a weight 1 kg to a height of -." m in - s. he inpt energy spply to the motor in onesecond is $" 2. #alclate
a) power of the motor
o/er = /or$ donetime ta$en
= m!h t
= ) )&4.0) 4
= #0 /att
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4etal ball
-" meter
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b) the efficiency of the motor
,fficiency = seful ener!y output x #001 ,ner!y input
= #0 x #00120
= (0 1
4/)5 A22RECIATING T6E IM2ORTANCE OF MA7IMISING T6E EFFICIENCY OF DE8ICES
1.