和
Hook
'
④
F K
Law
s
=
a
load
undertheir proportional
they
are
sprnysextend
-
long l 个mit
as
(
.
.
.
②
Vimit
of
F kx
=
Force
( Load )
proportional
in
to
P 9 q)
proportionaliy
Load
=
sprny
constant
extension
X
Csprry
dangd
Λ
1
—
extension
wiu
o
return o
↑
F kx
=
>
0
,
Extenson
Zero
not
2
1
(a) State Hooke’s law.
书本定义
..........................................................................................................................................
For
Examiner’s
Use
..................................................................................................................................... [1]
(b) Fig. 1.1 shows a graph of the stretching force F acting on a spring against the extension
x of the spring.
250
200
F /N
≈
150
.
k
100
F
=
,
50
F
F 5
K第
。
“
→
50
0
☆
β
0
10
20
30
40
50
60 70
x / mm
80
Fig. 1.1
(i)
State the features of the graph that show that the spring obeys Hooke’s law.
的
sraight
..................................................................................................................................'
Ume
pass throph orj
的
a
.............................................................................................................................
[1]
(ii)
Calculate k, the force per unit extension of the spring.
F
x
=
=
50
N
romm
2 cm
Ncm
25
k = ...................................................
[3]
© UCLES 2013
0625/31/O/N/13
.
3
(iii)
The limit of proportionality of the spring is reached at an extension of 50 mm.
Continue the graph in Fig. 1.1 to suggest how the spring behaves when the
stretching force is increased to values above 125 N.
[1]
(iv)
For
Examiner’s
Use
Another spring has a smaller value of k. This spring obeys Hooke’s law for
extensions up to 80 mm.
⼀
On the grid of Fig. 1.1, draw a possible line of the variation of F with x for this
spring.
[1]
[Total: 7]
© UCLES 2013
0625/31/O/N/13
[Turn over
4
3
Fig. 3.1 shows part of the extension-load graph for a spring.
0
extension / cm
3.0
k
B
步
=
lam
4~
=
A
0
0
12.0
load / N
3
Fig. 3.1
2 →
The spring obeys Hooke’s law between points A and B.
(a) (i)
(ii)
12
→
8
N
.
On Fig. 3.1, complete the graph between A and B.
[1]
State the name of point B.
proportionality
limve
.......................................................................................................................................[1]
of
.
(b) The average value of the load between A and B is 6.0 N.
Calculate the work done in extending the spring from A to B.
w
=
Fxd
=
6
3
×
=
18 J
↓
~
0
.
03 m
J
V
18
0
work done = ...........................................................[2]
.
© UCLES 2015
0625/33/O/N/15
5
原⻓
(c) The spring has an unstretched length of 4.0 cm.
—
An object is hung on the spring and the spring length increases from 4.0 cm to 6.0 cm.
⼀
(i)
Calculate the mass of the object.
F kx
=
m=
↴
时
(ii)
X
0
O
等
4
×
=
2 Cm
三
—
-
10118
=
0
.
8
.
0 8 时
mass = ...........................................................[3]
0
.
,
The object is immersed in a liquid but remains suspended from the spring.
The liquid exerts an upward force on the object and the length of the spring decreases
to 5.0 cm.
θ
X
e
=
5
-
4
=
1
cm
Calculate the upward force exerted on the object by the liquid.
F
⑤ε
© UCLES 2015
F
=
kx
=
4
4×1
=
4N
.
V
upward force = ...........................................................[2]
,
F
=
[Total: 9]
4r ↓
mg
0625/33/O/N/15
-
F
=
4
[Turn over
12
2
.
21
Homenork
3
(a)
定义
←
...................................................................................................................................................
...............................................................................................................................................[1]
(b)
A force of 120 N causes an extension of 64 mm.
⼀
(i)
On Fig. 2.1, draw the force-extension graph for the spring for loads up to 120 N.
[1]
150
force / N
(
0
,
0
)
100
(
64
,
20
straipht
50
)
lime
.
0
0
20
40
60
80
extension / mm
Fig. 2.1
(ii)
Calculate the spring constant k of the spring.
1
/
9
N
75
N
.
mm
/
18
cm
k = ...........................................................[2]
.
(c) A student makes a spring balance using the spring in (b). The maximum reading of this
balance is 150 N.
The student tests his balance with a known weight of 140 N. He observes that the reading of
the balance is not 140 N.
Suggest and explain why the reading is not 140 N.
oan
Hooke s Law
' t
Above 2 or ,
...................................................................................................................................................
the
obey
'
spriy
has been
The limie
...................................................................................................................................................
of
proportonaliy
spriy
if
exceeded
.
...............................................................................................................................................[2]
.
[Total: 6]
© UCLES 2017
0625/41/O/N/17
[Turn over
Moment
Amomentis the
.
of
the
uriing
Moment
(
effat
measuwe
on
a
body
.
Forceperpendin
a
=
⼀
( N)
Nm )
dis
tance
from
pivot
(
Increase
Momnt
Examples
Equilibrum
F ↑
,
d
)
↑
⾃⼰补充
:
:
:
m
①
no
net
②
no
net
~
force
moment
clockwisemoment =
(
Anti - dockwise moment)
id
O
d
=
√
→
O
↓
~
的
~
…
∠
×↑
dasdB
M
↓
>
=
F
∴
FA < FB
.
⼀
⻓
∝
↓
⼀
√
↓
l
*
⼤
⼤
×
D
√
√
×
√
。
wxa
=
Exb
.
☆
妥
是
⑥
参
λ
步
弯
⇌
⾔
⑧
羹
;
—
Pivot
⼀
⼀
⼀
cmscaler
1
=
2
KN
µ=
Fxd
=
1)
×
103
1
×
.
8
20
6 kN
20
=
kNm
kNm
×
控
Uvenicle
kN
=
Fxd
=
⇌
24 k
~
wilu oe
范围 T
1900
=
7
o
~
F
4 KN
m
≥
20
tips
×
10
×
1 25
.
24 kNm
krm
the vehicle
over
.
resultant ( net
moment
Resultontl
fora
nee
Tobalance
240
×
1
.
2
290
∞
F× 3
F
.
2
290
=
×
wm
3
.
2
F
=
90
N
the
weght
Scalar d
↓
Vevtors
.
speed
timed
onlyhas magniendersines
has
Diagram
1
2
3
.
.
mass
:
magnitude
both
,
,
a
diration
:
force
velocity
,
:
确定 scale
画精确
,1
,
cm
= 2
N
lov
or
补充完整图形 ( ⻆形
⻔呈直⻆
1 …
品壁
or
F
…
年⾏回也形 )
可以计算
F 2+
4 ==
=
θ
,
(2
cm
)
θ= tam
"
(
丢)
=
…
…
⻔呈⾮直⻆
F
FX
2
θ
「混是可以直接测量的
—
-
→
F
,
。
Momenum
It
.
'
(
product
the
s
Veor )
of
p
principle of
,
The
total
.
Imprlse
the
'
Zt s
i计
a⽐此
Impulse
F
=
=
.
momentum
doesn
and
constent
hteraction
:
bodies
between
MBUB
produe
=
MAVA
fore
of
+
MBVB
time
ad
whuh
for
moq
哔 笔
“
=
p
=
Impulse
=
=
F
=
器
F
=
Fot
c
琵
t
Such
ollisions
Fot
mv
=
(
'
。
a=
ma
T
of
s
momentum
of
relocity
o
mo
conservation
charfebecause
MAUA
=
masi
( mv )
as
)
①
)
⑬
⑤
UB
UA
A
T
MBUB
↑
MAUA
=
⑤
)
MAUA
静⽌
=
=
O
)
MAVA
MAVA
+ MBVB
+
MBVB
MAV
UA
?
VB ?
V
?
MB V
②
⼦弹⽊块
①
A
→
国
。
VA VB
=
MAUA
MA
+
V
晶≈
VAB
MBUB
MaL +MB
)
VAB
=
⼀
0
↑
。
0
—MA
Tpivot
= Fxd
UB : Fxd =
∴Ʃ M
=
=500 ×
400 X
1
.
.
2
1-
2
=
120 Nm
=
600 wm anti
480
120
wm
wm
⇌
xlw
+ 400
500 x
2
=
1080
Nm
→
Ʃ
F
=
500
-
400
⼀
100
N
clok
wise
dockwise
anticlockwise
M
-
MA 60001kg
=
UA
=
mB
=
0
5 mls
5000
UB
:
kg
0 m
/s
⼀
⼀
PA
=
MO
=
6000
×
5
30000
=
)
kgms
→
I
=
oP
=
27000
—
1k 8
m
/s
ct
I
F
=
I
/t
.
6
s
27000 kgmls
=
%
=2700
0
=
.
6
=
45000
~
momentwm
①
truck A
for
loss
truck
B
by
gam
⼀
momentum
=
P
∴
=
30000
27000
27000 kgmls
3000 的 m / s
=
Mo √
=
∴ cV
-
=
=
30
品品
0
=
.
5
mls
mBV
maVa +
MAUA
mBUB
+
6000
×
5
=
+
5000
×
0
6000
=
UA
×
UA
+
27000
30000 27
—
-
=
6000
=
0
.
5
mls
but
V
=
=
magniende
has
magniende
p
=
mo
,
358
1200
.
magniude
and
p
mls
'waton
hasbothd
it
=
mv
42 × 104 8 mls
k
.
.
.
Becanse
m
only
scalor
Vrtor
dwertonsand
both
has
Vertor
=±
e
mv
±α
=
=
=
⼀
2
.
5
35
×
×
12002
107 J
⼀
Constant
Increases
chemical
,
Iful energy hansfomelto
Kmetuenergy
.
P
=
MA
-
VA
=
2000 ×
18
36000
kgmls
⼀
36000
-
21000
gmlS
150001
15000 kgmls
—
0
t
=
0
.
2s
.
⼀
F
=
⺠ 瓷
=
5
:
器
75000
~
→
e
⼀
Increase tbme wll
of
chafe
mometum
of
smaler
forces
he
deweanrae
do
.
on
people 1
less
njury
Mh
=
0
.
15
g
mls
h
8
=
t
0
op
=
mov
=
=
mV
0
.
15
-
×
8
1
=
I
/
0t
=
1
0
.
mu
1
F
=
Y
0
.
.
.
21话 mls
2
kgmls
0015
SoON
0015s
.
ati
0.
⼀
⼀
k
p
√
-
e
=
=
Ʃ
mf
mu
e
×
×
×
×
×
√
×
×
Same
—
ar
p
mv
=
α
k e
-
Ʃ
=
104
=
mv
^=±×
×
Care
×
P
:
k
-
mv
=
102
50000
=
√
1000
104
=
~
e
=
Ʃ
m
' =±× 500
=
2x
(
O
× 20
5
⼀
→
U
=
0
~
MAUAt MAUB
2
×
12 +
3
× 0
V
=
MA
MB) V
+
5 V
=
/5
24
=
√
=
4 8
.
/s
m
~
,
→
…
…
⼀
√
√
ε
.竖直⽅向
F
⼀⽔平向右
=
0
。
∞
√
o
.
1
118
.
918
□
boomls
MAUA
0
.
U
+
1
×
MB
=
UB
600 +
0
√=
0
( MA
=
= (
1
.
9
号
+
+
=
MB
)V
o. V
30
~
1s
Power
Worka
W
,
unit
CE
=
Jonle
:
(J
)
↓
W=
-
Fxd
( Nm )
Power
P
is
哇
=
essure
-
Pressure
P
.
In
the
_
op
done
Unit
the
is
-
fore per
:
Wa
点)
-
,
unit
mr
area
unit
:
Pr
N
=
:
work
→∞
liqwid
P
of
.
FYA
=
rate
pgh
ogoh
As the
the
depth
pressuwe
lquid
/
=
m
的 creases
cansed by
mcreases
.
.
Measuriy
pressume
:
⑩
uawomeen
,
,
mercury
Barometer
latm
=
760
mm
15
g
.
A
戴
个州
Vacuum
…
length
measured
vacuum
stay the same
760mmHg-15mmHg=745mmHg
745mmHg
p1V1=p2V2
15mmHg×12cm³=p2×4cm³
p2=45mmHg
760mmHg-45mmHg=715mmHg
715mmHg
p=ρgh=840kg/m³×10N/kg×0.36m=3024Pa
3024Pa
10^5Pa
The height of the liquid level on the left is not as low.
the two liquid levels higher
p=ρgh=1020kg/m³×10N/kg×15m=153000Pa
153000Pa
153000Pa+10^5Pa=253000Pa
253000Pa
F=pA=253000Pa×1.2m×0.8m=242880N
242880N
moment of force changes
there is a pressure inside box
F=pA=5×10^5Pa×6.5×10^-4m²=325N
F1d1=F2d2
325N×7cm=F2×24cm
F2=94.8N