Tutorial 9A
①
7522N
25kg
Fr -12N
-
<
>
5m
a)
W=
Fs
(-12/15)
=
-60J
=
b)
W
Fs
=
( 22715 )
=
110J
=
c) Wd ,
110-60
=
50J
=
d)
Wd
I
=
mv2
tgmuz
-
I
50J
4
1,125)
=
2
-1-2
=
:
112511012
-
v2
=
v
v2
v
2ms
=
-1
Tj 22N
②
25kg
12N
,
Wd
(
=
I
=
I
1. 94
v
40°) / 5)
cos
( 12)( 5)
-
24.3J
=
24.3
22
=
=
125) v2
-
1,12571012
v2
1.39ms -1
u=
③
v=
150119
a)
increase
KE
=
=
=
b)
increase
:
KE
0ms
-1
4ms
gmv2 tgmu
"
≥
-
≤ ( 50 ) (4)
2-
0
400J
=
decrease
decrease
GPF
=
GPE
400J
u :O
U= 0ms
-85kg
m
-
H
"
2m
_
¥
h= 3M
130°
-
a)
GPE
H
mgh
=
2
=
(85/110)/3 )
=
sin
Length
2550J
=
7.14
=
wd
1mV
=
{
≥
_
3
2550
I
60
V
( 85 > v2
≥
)( 0 )
1,185
-
Fr=80°ᵗ
{ m(
'
_
2000N
=
200m
increase
7.75ms
Fp
,
2
v2
=
=
2
mu
10
1
=
4
-
3.14M
=
b)
4m
=
30°
v2
-
KE
Wdp
=
U2 )
Fs
=
Wd resistance
-
,
-
Fs
800
c)
1*(160/0) (
85+35
m=
120kg
=
800
GPE
v2
=
-
U2)
-
-
( 800 )( 200)
240000
=
U2
-
( 2000 )( 200 )
300J
=
3600J
=
3600
( v2
(12011/0) (3)
=
V2 U2 )
( 120 ) V2
1
=
When
0
-
✓
<
30ms
-1
v2
,
<
900m ,
-
1
2
60
V
v2
=
300+42
7.75ms
=
'
'
U2
U
8
9)
v
h
50kg
=
m
9.9ms
=
b-
=
,
ha
_
'
m
100M
=
increase
{
:
( 50 )( 9. 9)
KE
2-
Wd gravity
=
≥
1,150 ) (
O)
2450.25
=
100hr
=
Fr
b)
The
velocity
"
40kg
m=
9
high
is
"
%
=
=
Wd resistance
-
( 501110 ) / 5)
2500
u= 0ms
=
=
=
b)
when
there is
wd gravity
mgh
800
S
=
=
=
=
has
-1
2m
V
GPE loss
it
"
-
a)
Mgh
( 40 ) ( IO )
800J
no
KE
,
Wd resistance
Fs
( 112)s
7.14m
( 100 )
100 Fr
49.75
,
¥
Fr
0.498N
therefore
.
-
-
(2)
=
0ms
-1
a
small
resistance
.
the
u
900
<
600
<
L
±
value
24.5ms
should
_
'
be
less
than
24.5ms
-
'
\
Tutorial
"
"
6
9B
v.
I
✓
pug
,
8ms
=
h=
20°
,
h
7.10°
3.5M
\
§( 3) (8)
RE
=
20°
3.5 sin
1.20M
=
Decrease KE
tgmv
{ v2
I
2-
(3) (a) 2
I
GPF
decrease
=
(3) (
=
1.542
96
=
U2
)(I 2)
=
v
6.33ms
=
_
'
②
ztmv
might
≥
=
v2
( IO ) ( 1.43 )
=
I
v
③
m
57g
=
U
-1
0.057kg
=
-1
150 kmh
=
5.35ms
=
180000
=
50ms
=
60×60
GPE
a)
(0-0577110)/1 )
=
0.57J
=
b)
GPF
increase
0-57
V2
-
-
_
2500.57
=
v
-150ms
=
V2
0.57
=
v2
u
KE
1¥05 7) (5012-1*7)
=
502
decrease
=
-1
"
0ms
-
④
h= 0.2m
_
I
"
2.5m
increase
↳ HNVZ
KE
t.mu
_
decrease
=
=
2
↳ V2
v
(10110-2)
=
=
4
=
±
'
V
2ms
-1
v=0
'
⑤
GPE
nigh
'
2.5M
h
=
-
Ms
2.5
'
sin
,
730°
=
Decrease
RE
1- m/ v2
Increase
=
nigh
=
2
I v2
=
I
v2
V
=
=
( 10711.25 )
25
# MS
-1
1.25M
GPE
=
=
v2
-
40
=
4
10
36
-
=
"
'
increase
oms
30°
-1
"
'
=
=
GPE
increase
nigh
( IO )
(0-208)
4.16
2.04ms
"
1.2
0
-
sin
/ 0°
208m
10
1
Mm
GPE
lomh
RE
increase
)
increase
{
=
{
{
=
b)
≤
KE
MV
mv
£Mv2 -13mV
I
J
'
energy
10mn
=
'
10 Mhsino
Msin ⊖
-
20h ( m
=
{ MVZ
-
10h1m
=
ball
GPE
-
10 Mh
=
v2
0J
decrease
=
( Mtm )
v2
'
increase GPE
t
lomhsino
I
-
2
ME crate
-1
Lmu
'
mv
10 Mhsin
Mv 't
increase
'
J
mechanical
in
increase
=
=
m( 10th
=
iii
1
mgh
=
=
)
-
In
=
ii
0ms
=
u
⊖
I
a) it
u
-
Msino
)
)
( mtm )
v
M
20h1m
=
-
Masino)
(Mtm)
-0.2kg
-
A
11
45m
.
•
,
3.6m
,
c
a) it
KE
increase
≤ mv2
0 / v2
v2
£10
2) V2
.
It has
c)
new
i
m=
)
Iz
0.2×2
lgmv2
lo 4) V2
-
2
v2
v2
✓
0
±
=
=
'
(10113-61-2-45)
/ 1ms
-1
because
.
the
sculpture
is
slippery
-4kg
Mgh
10 4) ( IO ) / 2.45 )
-
49
±7
Ms
ii
)
£(
a.
4) v2
0.2
=
V2
v2
9.8
=
-
121
=
=
=
-
0
=
7ms
12.1
friction
no
-
=
✓
±
10 2)
=
v2
b)
=
=
0 / V2
4.9
49
=
✓
)
-
=
.
Ii
10 2) ( IO ) / 2.45 )
=
-
GPE
Mgh
=
( O 2) V2
±
decrease
=
V
-1
(0.41110113-61-2-45)
24.2
=
=
=
121
11ms
-1
I
Tutorial
9C
a)
KE
increase
GPE
increase
+
1mV
'
mgh
+
Wd
=
Fs
=
2
{ (60/110)
2-11-60 )( 10)( 7)
4200
-
=
3000
+
=
S
b)
ME lost
GPE
=
ME
lost
24m
=
3000
-
1200J
=
c)
-50s
KE
-
4200
=
( -50 )s
changed
is
R
heat
into
energy
.
I
,
3M
②
U
Fr
pug
v
3.25m,
=
5ms
=
'
'
50N
h
) 30°
F
=
=
-
-1
'
3 sin 30°
1.5m
10TH
f-
100
=
b- ON
=
R
100
=
MR
=
86.6M
=
increase
≤ MVZ
{
30°
COS
86.6 N
=
F
30°
sin
RE
in
{ Mui mgh
-
( IO ) (5)
"
_
=
work
(D) (10/11.5)
-
77.81
ME
=
ME
GPE
-
lost
RE
+
mgh
=
240
=
1000m,
v2
v
[
+
u=0
200.0m
,
5ms
-
I
295
¥000
=
180 /
216
02+240)( 1000)
=
increase
{
2
-
42
=
{ ( 12 ) / 6)
-
-
*
v2
-
2
24J
=
80kg
≤ Mr
(12/110)/2)
=
4
-
(5)
KE
+
2
_
ms
increase
-1
Pf
£180) (20-000)
=
-
Wd
(80/110112000)
-
2399000
R
=
=
=
-
-
C-
=
=
M
3
done
Fs
=
≤ ( IO ) ( 3.25)
2-
in PE
increase
-1
2000k
2000k
1200N
=
0
.
86.6M)(
259.8
299
3)
µ
Increase
b-
1
KE
(1) ( IO ) ( 3
≥
(1) ( 10.5 )
PE
Increase
+
-
Wd
=
20+5)
sin
=
-
3F
,
3F
f-
⑥
b- 0g
m=
u
S
0.050kg
=
PE
20m
=
10m
=
Ums
=
-1
0.5N
:
Increase
-
( 0.05 )s
-
=
S
Fr
Wd
=
(-0.051/10112)
U
KE
PE
Increase
+
lymu 't ymgh
0.55
gh
-1
Wd
=
-0.5ms
=
2
0.542
=
0.5
us
10.5
=
0
=
.
b- U2
yz
=
21
10.5
:S
8
m
50ms
=
s
0459kg
.
'
0.3N
=
Increase
KE
Increase
+
(0-0459) v2
1
0
m
160m
=
Fr
21
=
-
)
42
-1
5m
=
Sc
µ
=
45.9g
=
V
a)
1.71N
2M
=
s
5.14
0 -05N
=
Increase
7
=
0
=
Fr
=
-
PE
(0.0459/150)
0.023 v2
V2
V
KE
=
1
=
=
=
≥
( 0.0459) ( 22.5 )
2
=
KE
11
.
=
Increase
3.62
62
8J
after
=
-
'
-
,
,
b)
Wd
=
I
11.62-8
3.62J
KE
+
PE
Increase
(0-04591110)
-
s
(3-0.459)
2.
=
s
s
=
=
-3s
3.62
=
541s
=
Wd
3.62
1.42m
11.68
b- 07.8
22.5
Ms
-1
(0-0459) ( lo ) (5)
=
C- 0.3 )( 160)
Tutorial
I
9D
1 ( 2000 ) ( 2012
a)
!
_
2
Wd
P
300000
=
t
8
37.5kW
=
2
b- tan
p
f-
P
400k
V
3
=
t
=
V
-1
20ms
=
140kW
10000N
V
F
=
( 10000)✓
=
14ms
=
200g
_
8ms
v
Wd
20000
=
2ms
=
Fv
=
=
m=
U
20000
=
F
V
4
400K
=
=
140k
5000kg
=
P
p
'
-
0.2kg
=
'
"
3s
=
1
( 0.2 )( 8)
2
! / 0.2) / 2)
_
,
=
p
-
300 KJ
W
=
2000 )( 10 )
300000J
=
=
b)
(
6J
-6
=
'
2W
=
3
5
P
V
p
DF
=
120kW
'
=
10ms
=
FV
=
120kW
-
=
12000N
10
f-
=
=
•
:
DF
-
12000
Fr
-
Resistance
FR
force
=
12000N
=
Wd
Rr
"
%=
6
"
"
,
✓
ygg
FP
"
15ms
=
[ 100
18000N
f-
18000 sin 10°
=
3125.67
=
Fp
2000-13125.67
=
=
p
5125.67
FV
=
(5/25.67)/15 )
=
76885W
=
U
7
't
}
2778 ;
.
-
I
12mi
=
t
)l0°
10ms
=
v
16007
"
"
60s
=
16000N
v
utat
=
12
2
10
=
60A
=
a
-2
0.0333ms
=
F
a(60 )
+
Fn
=
Fr
-
160010.0333 )
P
=
2778.37
-
11
2831-65
P
=
Il
P
31148.15
=
31.1kW
D=
"d
>
8
p
prob
v
=
75kW
25ms
=
-
'
263£
, go
"
P
75k
f-
3000
-
1=125 )
=
3000N
=
263
12000N
Fu
=
1670.1
-
1066.9
a
12009
=
-9
9
1200
=
=
0.889
Ms
-2
16001g
m=
4=0
Er
2400N
:
D=
48kW
a)
P
48000
F
=
=
=
FV
Fu
48000
N
V
b)
F
2400--16009
-
48000
2406=16009
-
v
a
c)
Because
=
( 3,0
there
is
-
1.
also
5)
Md
the
-1
heat
lost
and
the spread is
increasing
.