1.

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.

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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