a man weighing 60 kg lifts a body of mass 15kg to top of a building 10m high in 3 minutes His efficiency is a) 20%b)10%c) 30% d) 40% plz answer fast...
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Answered by
189
Input power is the power produced by man to reach the top of the building along with the body
Input power = (M + m)gh/t
Output power is the power required to take the body to top of the building
Output power = mgh/t
Effeciency = (Output power / Input power) × 100
= [ (mgh/t) / ((M + m)gh/t)] × 100
= [m / (M + m)] × 100
= [15 kg / (60 kg + 15 kg)] × 100
= 20%
Input power = (M + m)gh/t
Output power is the power required to take the body to top of the building
Output power = mgh/t
Effeciency = (Output power / Input power) × 100
= [ (mgh/t) / ((M + m)gh/t)] × 100
= [m / (M + m)] × 100
= [15 kg / (60 kg + 15 kg)] × 100
= 20%
mallikarjun5:
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Answered by
42
Total work = Work done when man moves to top of building + Work done when the object is moved to top of building
Power = Work / Time
3 minutes =180 seconds
Power Input = (60×10×9.8)/180 + (15×10×9.8)/180
=98/3 + 49/6
=245/6
Power output = Work done to move the object / Time
=49/6
Efficiency = (Power input/Power Output)×100%
=20%
Power = Work / Time
3 minutes =180 seconds
Power Input = (60×10×9.8)/180 + (15×10×9.8)/180
=98/3 + 49/6
=245/6
Power output = Work done to move the object / Time
=49/6
Efficiency = (Power input/Power Output)×100%
=20%
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