Fig. 1 shows a lifting test at a scrapyard. An electromagnet is supplied by a 12 V battery. The current is 2.5 A and the crane lifts a 60 kg bundle of steel...
Question 1 Report
Fig. 1 shows a lifting test at a scrapyard. An electromagnet is supplied by a 12 V battery. The current is 2.5 A and the crane lifts a 60 kg bundle of steel for 8.0 s. Assume all electrical energy supplied is transferred to the lifting motor.
(a) Calculate the electrical power supplied to the electromagnet.
Use the equation: power = potential difference × current. [2]
(b) Calculate the electrical energy supplied in 8.0 s.
Use the equation: energy = power × time. [2]
(c) What is the magnetic material in the bundle? [1]
(d) Give one reason why a real crane transfers less useful energy than the value calculated in part (b). [1]
Answer Details
Electrical power is the rate of energy transfer.
- (a) \[P=VI=12\ \text{V}\times2.5\ \text{A}=30\ \text{W}\] [2]
- (b) \[E=Pt=30\ \text{W}\times8.0\ \text{s}=240\ \text{J}\] [2]
- (c) The magnetic material is steel. [1]
- (d) In a real crane, some energy is dissipated as heat or sound, or is used moving other parts of the crane. [1]
Download The App On Google Playstore
Everything you need to excel in your exams
