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A machine is driven by a 100 kg mass that falls 8m in 4*0s. it lifts a load of mass 500 kg vertically upwards. Taking g=10ms^(-2). Calculate: (a) the force exerted by the falling mass. (b) the work done by the falling mass in its displacement by 8*0m. (c) the power input to the machine. (d) the power output of the machine if its efficiency is 60%, and (e) the work done by the machine is 4*0s. |
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Answer» Solution :Given: `m=100kg,d=8*0m,t=4*0s,L=500kg` (a) Force exerted by the falling mass=mg `=100xx10=1000N` (b) Work done by the falling mass =force exerted by the falling mass`xx`displacement `=mgxxd` `=1000xx8*0=8000J` (c) Power input`=("work done on machine")/("time")` `=(8000J)/(4*0s)=2000W` (d) Given, EFFICIENCY=60%=`0*6` power input=? SINCE efficiency=`("power input")/("power input")` `therefore`Power output=power input`xx`efficiency `=2000Wxx0*6=1200W` (E) Work done by the machine is `4*0`s =power output `xx` time `=1200Wxx4*0s=4800J`. |
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