A machine of velocity ratio 6 requires an effort of 400 N to raise a load of 800 N through I m. Find the efficiency of the machine
Answer Details
The efficiency of a machine is the ratio of the output work to the input work. In this case, the output work is the work done on the load and the input work is the work done by the effort. Given that the velocity ratio of the machine is 6, this means that for every 6 meters of rope pulled by the machine, the load is raised by 1 meter. Therefore, the distance moved by the effort is 6 times the distance moved by the load. Hence, the input work done by the effort is: Input work = effort × distance moved by effort Input work = 400 N × 6 m Input work = 2400 J The work done on the load is the output work. Since the load is raised through a height of 1 m, the output work done on the load is: Output work = load × distance moved by load Output work = 800 N × 1 m Output work = 800 J Therefore, the efficiency of the machine is: Efficiency = output work / input work Efficiency = 800 J / 2400 J Efficiency = 1/3 Efficiency = 0.333 Converting to a percentage, we get: Efficiency = 0.333 × 100% Efficiency = 33.3% Therefore, the answer is 33.3%.