Question

If a load f 1200 kg is lifted by an effort of 300 kg by a simple machine having velocity ratio of 5 ,then its efficiency is ____

Answer 1

The answer is 80%.


Explanation:

mechanical advantage

= load/effort
= 1200/300
=400

Given: velocity ratio= 5

Therefore,


Efficiency
= mechanical advantage/ velocity ratio
=400/5
=80

As efficiency is always expressed in %,
Therefore, efficiency= 80%

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%