A block and tackle is used to raise a load of 250N through a vertical distance of 30m. What is the efficiency of the system if the work done against friction is 1500J?
- 62.5%
- 73.3%
-
83.3%
- 94.3%
The correct answer is: C
Explanation
Y = \(\frac{work output}{work input} \times \frac{100}{1}\)
Y = \(\frac{250 \times 30}{250 \times 30 + 1500} \times \frac{100}{1}\)
There is an explanation video available .