A concrete slab with a mass of 3 tons was evenly lifted to a height of 5 m, having performed a work of 160 kJ for this. What work is spent on overcoming the frictional force? what is the efficiency of the mechanism?
A1 = 160000J
A2 is equal to theoretical work in the absence of friction and is equal to potential energy
A2 = mgh = 30000 * 5 = 150000j
The work of the friction force is 160-150 = 10kJ
The efficiency is 150/160 = 94%
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