CIE iGCSE Co-Ordinated Science P1.3 Mass and weight Exam Style Questions Paper 4
Question
Fig. 3.1 shows a forklift truck lifting a crate.
(a) The crate has a mass of 140 kg.
(i) Calculate the weight of the crate.
The gravitational field strength, g, is 10 N/kg.
▶️Answer/Explanation
Answer: 1400 N
Explanation:
Weight is calculated using the formula \( W = m \times g \), where \( m \) is the mass and \( g \) is the gravitational field strength.
Given: \( m = 140 \, \text{kg} \), \( g = 10 \, \text{N/kg} \).
Substitute the values into the formula:
\( W = 140 \times 10 = 1400 \, \text{N} \).
(ii) Calculate the work done on the crate when it is lifted through a height of 1.5 m.
State the unit for your answer.
▶️Answer/Explanation
Answer: 2100 J (Joules)
Explanation:
Work done is calculated using the formula \( \text{Work} = \text{Force} \times \text{Distance} \). Here, the force is the weight of the crate, and the distance is the height lifted.
From part (i), the weight \( W = 1400 \, \text{N} \), and the height \( d = 1.5 \, \text{m} \).
Substitute the values into the formula:
\( \text{Work} = 1400 \times 1.5 = 2100 \, \text{J} \).
The unit for work is Joules (J).
(b) The forklift truck uses an electric motor to lift the crate.
Fig. 3.2 shows the circuit that includes the electric motor.
The voltmeter displays a reading of 0.50 V.
(i) Show that the potential difference (p.d.) across the motor is 11.5 V.
▶️Answer/Explanation
Answer: \( 12 \, \text{V} – 0.50 \, \text{V} = 11.5 \, \text{V} \).
Explanation:
The total voltage in the circuit is 12 V, and the voltmeter reads 0.50 V. The potential difference across the motor is the total voltage minus the voltmeter reading:
\( \text{p.d. across motor} = 12 – 0.50 = 11.5 \, \text{V} \).
(ii) The current in the circuit is 9.20 A.
Calculate the resistance of the motor.
▶️Answer/Explanation
Answer: 1.25 Ω
Explanation:
Resistance is calculated using Ohm’s Law: \( R = \frac{V}{I} \), where \( V \) is the potential difference and \( I \) is the current.
From part (i), \( V = 11.5 \, \text{V} \), and \( I = 9.20 \, \text{A} \).
Substitute the values into the formula:
\( R = \frac{11.5}{9.20} = 1.25 \, \Omega \).