Question

If a load F 1200 kg is lifted an effort of 300 kg by a simple machine having velocity ratio of 5, then it's efficiency is -

Answer 1

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Answer 2

Answer:

80%

Step-by-step explanation:

The formula used for Mechanical Advantage is:

$\text{Mechanical Advantage}= \frac{\text{Load}}{\text{Effort}}$

= $\frac{1200}{300}$

= 400

Also, Velocity Ratio = 5

Efficiency is calculate by:

$\text{Efficiency}= \frac{\text{Mechanical Advantage}}{\text{velocity ratio}}$

= $\frac{400}{5}$

= 80

Hence, Efficiency = 80%