JAMB - Physics (2012 - No. 9)
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%
Explanation
Y = \(\frac{work output}{work input} \times \frac{100}{1}\)
Y = \(\frac{250 \times 30}{250 \times 30 + 1500} \times \frac{100}{1}\)
Y = \(\frac{250 \times 30}{250 \times 30 + 1500} \times \frac{100}{1}\)
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