(19)音教,利畫 日四十月一十年巴己廣夏
▲在法國,一名失踪十B. 的土耳其男童發現被殺害。
▼東柏林數千人在街頭舉行反政府示威
▲蘇聯軍隊接受防纂訓練。
HINESE SE TEEZING AS 225 25 FEET-
▼一名母親緊抱兒子 - 接受警察的保,祂是在參觀米產重
▼中國韓愛萍在新加坡公開羽毛球賽準決賽中 ▲珠王比利出席世界杯抽簽的儀式。
以二比一擊敗對手。
Tona SEIKO
KO SEIKO
1990 中學會考預習專欄
明德出版社
MILL & DALE PRESS
報日僑華
一期星日一十月二十(九八九一)年八十七國民華中
****
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Solutions
1. Let T-tension in the cord
a-acceleration of the block mass of the block-10 kg
Apply Newton's 2nd aw
T- mg - ma
As shown in the force diagram, consider the forces act of the man R reaction exerting on the man.
Apply Newton's and law
R.
Exerc
put c
putu
put t
ma
1. 1
800 = 80(1.5)
R = 230 N
(Ans.)
15kg 10 kg
2.5
Physics (11)
C. Y. Mak
Exercise 11 Force and Acceleration (1)
(where necessary, take y = 10 ms-2)
1. A 10 kg block. is supported by a cord and pulled upwards with an acceleration 2 ms 2.
(a) What is the tension in the cord?
(b) After the block has been set in motion, the tension in the cord is roduced to 100 N. What sort of motion will the block perform?
(c) If the cord is now slackened completely, the block is observed to mave, up 5 meters further before .coming to rest. With what velocity was it travelling?
2.
In figure 1, a man of mass 80 kg stands on a platform of mass 40 kg. He pulls a rope which is fastened to the platform and runs over frictionless pulley on the ceiling. The platform is observed to move
upward with acceleration 1.5 ms Find
(a) the tension in the rope, and
Floor
Figure 2
If the system is now released,
(a) determine the acceleration of the
system,
(b) find the tension in the string,
(c) find the velocity with which the
15. kg block strikes the floor, (d) Find the time required for
the
15 kg block to reach the floor, and (e) find the maximum height reached by
the 10 kg block.
4:
An inclined, plane AB 100 m long, B is on the ground and A is 60 m above the ground." (as shown in figure 3) A block X at A is allowed.
slide down from rest. velocity-time graph is as shown in figure 4.
to
R
-?
(b) the reaction from the platform
exerting on the man.
25-
20-
the
(a) T (10)(10) = (10)(2)
7 = 120 N
Ans: The teasion in the cord is
120 N.
(b) If T = 100 N, then
100
(10)(10) = 10a
-2
a = 0 ms
Thus, the block will ascend with uniform velocity.
(Ans.)
(c) If the cord is slackened, then
TO hence,
0 (10)(10) = (10)(a)
-2 --10 15 (The negative sign indicates that the direction of acceleration is downwards)
Apply the formula v2 - u2 + 2as
0 = u2 + 2(-10)(5)
u = 10 ms
-1
211
Figure 3
800
205
Sko
-2
10kg
(a) Let the acceleration of the system
-2 be a ms tne string be TN: Apply Newton's 2nd law blocks seperately
$5kg 10kg
velocity of the 15 kg black
when it strikes the floor
--1
=.4 ms"
Then, the string slackens, assume the block moves h meters. farther before drops. down.
= 2gh
42 =:2(10)h
3. 1
h = 0.8 m
.. the maximum height reached by
the 10 kg block
(Ans.)
= 4 m +
0.8.
#. 4,8
tension in
the
to
the
4.
15(10) - T = 15a
(Ans.)
T (10)(10) = 10a
(1) + (2)
(1) (2)
(a) The linear acceleration of block X
- slope of the linear graph
-1
= 2 ms
50 = 25a
(b) Apply
-2
a 2 ms
(Ans.)
$
11/1
at2
100 = {2}t2
t = 10 s
(b) Substitute the value of a into (2);
we obtain
T = 120 N
{Ans.)
(c) Apply the formula v2 = u2 + 2as
(Ans.)
(Ans.)
(c) Let the velocity of block X at B be.
E
Money
A. Mc
Con
Def
Fur
ищу
Platform
Figure 1
velocity (in s
10-
9 10
Time In second
Figure 4
a
3. Two blocks of mass 10 kg and 15 kg
are connected by
light inextensible string which passed Over
smooth fixed pulley. Initially, the 10 kg block rests on the horizontal floor and the 15 kg is held 4 metres above the floor as shown in figure 2.
(a) What is the linear acceleration of
the block?
(b) Find the total time taken for the
block to reach 9.
(c) Find the velocity of the block at
8.
(d) Given that the mass of X is 2 kg, find the frictional force between the block and the inclined plane
AR.
Bar
Ecc
(Ans.)
(a) As shown in the force diagrams
T = tension in the rope
(b)
Weight of the man and the platform = (80 + 40) 9
= 120 g
Apply Newton's 2nd law
2T - 120 g = (80 + 40)(1.5) 2T (120) (10) = 180
- T - 690 N
(Ans.)
where s 4 m, u = 0 ms
a = 2 ms
-2
*** v2 = 0 + '2{2}(4)
-1
v = 4 ms
(Ans.)
Thus, the velocity with which the 15 kg block strikes the floor is
-1 4 ms
(d) Apply the formula v u + at
where v = 4 ms
+2
, u = 0 ms
2 mis
.", 4 = 0 + 2t
t = 25
{Ans.)
(e) As shown in the figure, let the velocity of the 10 kg block be when it reaches the level at A.
y ms-t
v2 = 2as
2(2)(100)
= 400
V = 20 ms
-1
(d) Let m mass of the block x
9
inclination of the plane AB
The net force acting on X in the direction parallel to the inclined plane is
mgsing - f
where f frictional force
.*. the net force is
mgsine f = ma
where a is the accleration of the block
.*. (2)(10)() - f = (2)(2)
f = 8 N
(Ans.)
De:
Pri