34. A crane is driven by a lifting motor The motor provides power that can lift a 350 kg object at 8.0m/s . It is observed that in fact the crane lifts a 300 kg concrete block through 25 m vertically in 5.0 s. a) Find the power input of the crane. b) Find the power output of the crane. c) Find the efficiency of the crane.

Question

34. A crane is driven by a lifting motor The motor provides power that can lift a 350 kg object at 8.0m/s . It is observed that in fact the crane lifts a 300 kg concrete block through 25 m vertically in 5.0 s. a) Find the power input of the crane. b) Find the power output of the crane. c) Find the efficiency of the crane.

Answer 1

**a) Power Input of the Crane:**The power input refers to the power the motor *could* provide based on its rated lifting capacity. We are told it can lift a 350 kg object at 8.0 m/s. Power is the rate of doing work, and in this case, the work is done against gravity.* **Work done (W) = Force (F) x Distance (d)*** **Force (F) = mass (m) x acceleration due to gravity (g)*** **Power (P) = Work/Time (t)** or, since distance/time = velocity (v), **Power (P) = Force x velocity**So, Power input = (350 kg) * (9.8 m/s²) * (8.0 m/s) = 27440 W or 27.44 kW**b) Power Output of the Crane:**The power output is the actual power delivered when lifting the 300 kg concrete block. We need to calculate the work done and divide by the time taken.* Force (F) = (300 kg) * (9.8 m/s²) = 2940 N* Distance (d) = 25 m* Time (t) = 5.0 s* Work done (W) = (2940 N) * (25 m) = 73500 J* Power output = Work/Time = 73500 J / 5.0 s = 14700 W or 14.7 kW**c) Efficiency of the Crane:**Efficiency is the ratio of useful power output to the total power input, expressed as a percentage.* Efficiency = (Power Output / Power Input) * 100%* Efficiency = (14700 W / 27440 W) * 100% = 53.6% (approximately)Therefore:* **a) Power input:** 27.44 kW* **b) Power output:** 14.7 kW* **c) Efficiency:** 53.6%

Question

Solution

Answer