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MODUL • Fizik TINGKATAN 5

Latihan / Exercise

1 Sebuah cerek elektrik digunakan untuk mendidihkan air. Cerek itu mempunyai kuasa 3 kW, dan diisi dengan 0.5 kg air paip
bersuhu 20 °C. Ia mengambil masa satu minit setengah untuk mendidih. Berapakah kecekapan cerek itu? [Muatan haba tentu
air = 4 200 J kg °C ]
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KBAT An electric kettle is used to boil some water. The kettle has a power rating of 3 kW, and is filled with 0.5 kg of tap water at 20 ºC.
It takes one and a half minutes to boil. How efficient is this kettle? [Specific heat capacity of water = 4 200 J kg ºC ]
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Penyelesaian
Solution
Kecekapan cerek elektrik P
Kuasa / Power = P = 3 000 W = output × 100%
input input input The efficiency of the kettle P input
E output = mcθ 1866.67 W
= 0.5 kg × 4 200 J kg ºC × (100 – 20)ºC = 3 000 W × 100%
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= 168 000 J = 62.22%
\ Kuasa / Power = P
output output output
168 000 J
= = 1866.67 W
90 s
UNIT 2
2 Rajah menunjukkan motor elektrik yang digunakan untuk mengangkat beban.
The diagram shows an electric motor used to lift a load.
UNIT 2
Motor elektrik / Electric motor
KBAT

V 6 V

Beban
Load A
(a) Nyatakan perubahan tenaga yang berlaku apabila suis ditutup.
State the energy transformations that take place when the switch is closed.
(b) Jisim beban ialah 0.8 kg dan dinaikkan setinggi 1.5 m dalam 4.0 saat. Bacaan ammeter dan voltmeter yang sepadan ialah
1.2 A dan 5.0 V. Kirakan
The mass of the load is 0.8 kg and is lifted to a height of 1.5 m in 4.0 seconds. The reading of the ammeter and voltmeter are
1.2 A and 5.0 V respectively. Calculate
(i) tenaga output pada motor. / useful energy output of the motor.
(ii) tenaga input pada motor. / energy input to the motor.
(iii) kecekapan motor. / the efficiency of the motor.
(c) Apabila beban dialihkan dan suis ditutup, apakah yang akan berlaku pada
When the load is removed and the switch is closed, what will happen to
(i) bacaan ammeter? / the reading of the ammeter?
(ii) kuasa motor? / the power of the motor?
Penyelesaian
Solution
(a) Tenaga elektrik ditukarkan ke tenaga keupayaan graviti (iii) Kecekapan = 12 J × 100%
Electrical energy is changed to to gravitational potential energy Efficiency 24 J
(b) (i) E output = mgh = 50%
= 0.8 kg × 10 m s × 1.5 m
-2
= 12 J (c) (i) Bertambah / Increase
(ii) E input = VIt (ii) Bertambah / Increase
= 5.0 V × 1.2 A × 4.0 s
= 24 J

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