With the help of the movable Block, having an efficiency of 0.4, lifts a load weighing 50 kg to a height of 5 m. Determine the force applied to the end of the cable.
Efficiency = A useful / A spent
And useful = mgh
A spent = FS = Fh2
Efficiency = mgh / Fh2
We express from this formula F
F = mgh / 2hkpd
F = 50 * 10 * 5/2 * 5 * 0.4 = 625H
Answer: F = 625 N.
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