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.

Answer 1

Given : Effort E = 50 Kgf = 50 × 10 N = 500 N, d_E = 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 d_L 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 d_L =`(500×10×0.6)/2500`

= 1.2 m.