Fig. 1 shows a harbour crane lifting a box of engineering parts from a quayside. The box has a mass of 240 kg and rises vertically through 6.0 m at constant...
Question 1 Report
Fig. 1 shows a harbour crane lifting a box of engineering parts from a quayside. The box has a mass of 240 kg and rises vertically through 6.0 m at constant velocity. The crane motor transfers 18 000 J of energy to the lifting cable. Students use the diagram to check the direction of forces and to decide whether the motor is efficient. Take gravitational field strength, g, as 10 N/kg. Air resistance on the box is very small. The arrow labelled X represents one force acting on the box.
(a) What is the name of the force represented by X? [1]
(b) Calculate the weight of the box. [2]
(c) Calculate the useful energy gained by the box. [2]
(d) Calculate the efficiency of the crane motor. [2]
(e) Which energy store of the box increases as it rises? [1]
Answer Details
The question tests forces, gravitational potential energy and efficiency.
- X is the tension, or pulling force, exerted by the cable on the box. [1]
- Weight is calculated using \(W = mg\):
\[W = 240\text{ kg} \times 10\text{ N/kg} = 2400\text{ N}\]Weight = 2400 N. [2] - At constant velocity, the useful energy gained is gravitational potential energy:
\[E = Wh = 2400\text{ N} \times 6.0\text{ m} = 14\,400\text{ J}\]Useful energy gained = 14 400 J. [2] - \[\text{efficiency} = \frac{\text{useful output energy}}{\text{input energy}}\times100 = \frac{14\,400}{18\,000}\times100 = 80\%\]Efficiency = 80%. [2]
- The box's gravitational potential energy store increases as it rises. [1]
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