A block and tackle of 5 pulleys is used to raise a load of 50 kgf steadily through a height of 20 m. the work done against friction is 2000 J. Calculate:
work done by the effort
efficiency of system
mechanical advantage
Asked by subhasray1968 | 15th Dec, 2017, 09:47: PM
Expert Answer:
Number of pulleys = 5
Load = 50 kgf= 50x10=500 N (g=10 m/s2)
work=load x hieght = 500 x20 = 10000 J
Energy lost to overcome friction = input work - output work
2000= input work - 10000
input work= 12000 J
Answered by Gajendra | 18th Dec, 2017, 01:12: PM
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