Page 22 - Hybrid PBD KSSM 2021 Tg 5 - Fizik

Fizik Tingkatan 5 Bab 1

Lif (ii) Lif itu turun ke bawah
Lift The lift goes down
F = ma
mg – R = ma
R = mg – ma
= 50 × 10 – 50 × 4
Mesin penimbang
Weighing machine a = 4 m s –2 R = 300 N
Bacaan penimbang
Reading of the weighing
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machine
mg = 300 N
(iii)
Nota Visual F =
Gerakan Lif R – mg = m × 0
Motion of Lift (halaju seragam, pecutan
sifar)
(constant velocity, zero
acceleration)
4 m s –1 R R = mg
= 50 × 10
= 500 N
mg Bacaan penimbang
Reading of the weighing machine
= 500 N

(f) Rajah di bawah menunjukkan satu beban 4 kg (i) R
disambungkan dengan seutas tali kepada sebuah
blok logam 6 kg yang diletakkan pada sebuah T
meja melalui satu takal licin. 6 kg
The diagram below shows a 4kg load connected by a rope to a 6 F R T
kg metal block placed on a table over a smooth pulley.
W
6 kg 4 kg

B
4 kg
(ii) Blok 6 kg / 6 kg block
T – F = 6a
(i) Lakarkan gambar rajah jasad bebas bagi troli R
dan pemberat. T – 10 = 6a
Sketch the free body diagram of the trolley and the load. T = 6a + 10 ....... (1)
W = berat blok logam Beban 4 kg / 4 kg load
the weight of the metal block B – T = 4a
R = tindak balas normal ke atas blok logam, 40 – T = 4a
normal reaction on the metal block T = 40 – 4a ..... (2)
F = daya geseran Menggantikan (1) ke dalam (2), / Replace (1) into (2)
R
friction force 6a + 10 = 40 – 4a
T = tegangan tali 10a = 30 –2
tension of the string a = 3.0 m s

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Bab 1.indd 10 11/19/21 4:05 PM