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An effort of 250N raises a load of 900N through 5m in a machine. If the effort moves through 25m, find:
a) The useful workdone in raising the load.
Useful workdone=load*load distance =900*5= 4500J
b) The workdone by the effort.
Workdone by the effort= effort*effort distance= 250*25= 6250J
c) The efficiency of the machine.
Efficiency= (work ouput/work input)*20= (4500/6250)*20 = 72%.
Moraa orina answered the question on April 9, 2018 at 16:49
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