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Question
A machine is driven by a 50 kg mass falling at a rate of 10.0 m in 5s. It lifts n load of 250 kgf. Taking the force of gravity on 1 kg mass as 10 N. Calculate the power input to the machine. If the efficiency of the machine is 60%, find the height to which the load is raised in 5s.
Solution
Given : Effort E = 50 Kgf = 50 × 10 N = 500 N, dE = 10.0 m, t = 5s.
L = 250 kgf = 250 × 10 N = 2500 N, efficiency = 60% = 0.6
Power input to the machine =`"Effort×Displacement of effort"/"Time"`
`=(500xx10.0)/5`
= 1000 W.
Let the load is raised to a height dL in 5s, then
Efficiency =`"Power output"/"Power input"=("L"×("d"_"L"//"t"))/("E"×("d"_"E"//"t"))`
or 0.6 = `(2500×"d"_"L")/(500×10)`
or dL =`(500×10×0.6)/2500`
= 1.2 m.
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