Question

If a load f1200kg is lifted by an effort of 300kg by a simple mechine having velocity ratio of 5, then its efficiency is

Answer 1

velocity ratio = 5

so, Voutput /Vinput = 5

Load , F = 1200 kg ; Effort, F' = 300 kg

As we know that

efficiency, n = F.Vout / F'.Vin

or, n = F/F' × Vout/Vin

or, n = 1200/300 × Vout/Vin

or, n = 4 × 1/5 = 4/5 = 0.8

So, efficiency is 80 % .

Answer 2

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Load (L)=1600 kg

Effort(E)= 400 kg

Velocity Ratio (VR)=5

Efficiency = ?

Firstly,

Machanical advantage (MA) = Load/Effort

= 1600/400

=4

Then,

Efficiency= MA/VR*100%

= 4/5*100%

=80%

Efficiency of simple machine is 80%.

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