Question

If a load F 1200 kg is liftd by an effort of 300 kg by a simple machine having velocity ratio of 5, then its fficiency is

Answer 1

Given: load=1200 kg, effort=300 kg, V.R.=5;
M.A.=load/effort
= 1200/300
= 4
Efficiency= M.A./V.R.X100%
= 4/5X100%
= 80%