Calculate the power of an engine required to lift 10^(5)Kg of coal per hour from a mone 360 m deep (Take g=10 m s^(-2))

Calculate the power of an engine required to lift 10^(5)Kg of coal per

1.	Calculate the power of an engine required to lift 10^(5)Kg of coal per hour from a mone 360 m deep (Take g=10 m s^(-2))
Answer» Solution :m=`10^(5)` KG ,g=10 m `s^(-2)` ,h=360 m. t=1 h=60`xx` 60 s=3600 s The work needed in LIFTING a mass m to a height h against the force due to gravity is `W=mgxxh=mgh` and Power P= `("Work done")/("Time taken")=(mgh)/(t)` `therefore P=(10^(5)xx10xx360)/(3600)-10^(5)W=100 kW` Note:In fact,the actuual power of the engine will be MUCH more than 100 kW because (i)some energy will GET wasted in overcoming the force of friction and (ii)the efficiency of the engine will always be LESS than 100%