A pulley system with VR = 4 is used to lift a load of 175 kgf through a vertical height of 15 m. The effort required is 50 kgf in the downward direction, (g = 10 N kg-1). [4] Calculate:
Work done by the effort.
Topic: Machines, levers and pulleys
Answer
Given: effort E = 50 kgf, g = 10 N kg⁻¹, d_E = 60 m (from part (i))
First convert the effort from kgf to newtons: E = 50 kgf = 50 × 10 = 500 N
Work done by the effort = effort × distance moved by the effort = 500 × 60 = 30 000 J
Work done by the effort = 30 000 J (30 kJ)
Two things must match up before you multiply. The force has to be in newtons, so convert the 50 kgf using g = 10 N kg⁻¹ — that is why the question hands you g. Leaving it as 50 gives 3 000, which is not in joules at all.
And the effort's work uses the EFFORT's distance, 60 m, not the load's 15 m. Pairing the effort force with the load distance is the other common slip; it gives 7 500 J, which is actually the useful work output, a different quantity.