Spring
Diagram 9.1.1 shows a comparison of springs P and Q. Both springs are identical but different length. Diagram 9.1.2 shows the state of the spring when it is attached to load of equal weight.
Rajah 9.1.1 menunjukkan perbandingan spring P
dan Q. Kedua-dua spring adalah serupa tetapi mempunyai panjang asal yang
berlainan. Rajah 9.1.2 menunjukkan
keadaan spring apabila diletakkan beban yang sama berat.
(a) (i) State Hooke’s law.
Nyatakan Hukum Hooke
[1 mark]
(ii) Based on Diagram 9.1.1 and Diagram 9.1.2, compare the forces applied on the spring and the extension if the spring and the spring constant P and Q. Hence relate the spring constant and the natural length of the spring.
Berdasarkan Rajah 9.1.1 dan 9.1.2,bandingkan daya yang dikenakan ke atas spring dan pemanjangan spring dan pemalar spring P dan Q. Seterusnya hubungkaitkan pemalar spring dengan panjang asal spring.
[5 marks]
(b) Spring A with unstrecthed length 10cm is extended by 5mm by an applied force of
2.0N.
Spring A dengan panjang asal 10cm telah memanjang sebanyak 5mm apabila dikenakan daya sebesar 2.0N
(i) calculate the spring constant?
Kirakan pemalar spring
[2 marks]
(ii) calculate the extention if a load of 250g was suspended from it.
Kirakan pemanjangan spring jika beban 250g digantung padanya.
[2 marks]
(c) A cradle as shown in Diagram 9.1.3 can hold a kid of a maximum weight of 60N.
Buaian dalam Rajah 9.1.13 bolah menampung berat maksimum kanak-kanak 60N.
| |||
Using the appropriate physics concepts, explain the modification required to the cradle so as to hold a kid of weight 250N. In your explaination, emphasis on the:
Menggunakan konsep fizik yang sesuai, terangkan pengubahsuaian yang perlu dilakukan ke atas buaian untuk menampung kanak-kanak seberat 250N. Dalam penerangan anda jelaskan dari segi:
(i) design of the cradle
rekabentuk buaian
(ii) material of the cradle
bahan yang digunakan untuk membuat buaian
(iii) the strength of the spring of the cradle
kekuatan spring buaian
(iv) arrangement of the spring
susunan spring buaian.
[10 marks]
Hooke’s law state that the
extension of the spring is directly proportional to the applied force
provided that elastic limit is not exceeded
|
||||||||||||
Force that acted on the
springs are the same
Both springs are extended
Spring Q is extended more
than spring P
Spring P has a bigger
spring constant than spring Q
The shorter the spring, the
smaller the spring constant
|
||||||||||||
k = F/x ,
= 2.0N / 0.005m
= 400Nm-1
|
||||||||||||
x = F/k
= 0.25N / 400Nm-1
= 6.25 x 10-4m
|
||||||||||||
|