Question

Calculate the power of an engine required the lift 105 Kg of coal per hour from a mine 360m deep. (Take g = 10 m/s²)

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Answer 1

$\text\green{Question}$

Calculate the power of an engine required the lift 105 Kg of coal per hour from a mine 360m deep. (Take g = 10 m/s²)

$\text\green{Given :}$

${\mapsto mass = 105kg }$

${\mapsto height = 360m}$

${\mapsto time = 1hour = 3600sec }$

${\mapsto g = 10m/s² }$

$\text\green{Pre-Knowledge}$

${\mapsto Power = \frac{Energy/Work}{Time}}$

${\mapsto Power = \frac{mgh}{t}}$

${\mapsto Power = \frac{105 × 10 × 360}{3600}}$

$\underline\pink{\mapsto Power \: = 10⁵ \: W \: = \: 100kW}$

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Power - Energy relationship and Potential energy has been used. We see that we need to find the Power of the engine to do the process. Firstly we need to find the Energy required by the engine for lifting. Since, here the engine lifts the coal of some mass to some height, then the form of energy will be Potential Energy. Firstly we shall find the Potential Energy which is work done and then we shall find Time.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{Potential\;Energy\;=\;\bf{m\:g\:h}}}}$

$\\\;\boxed{\sf{\pink{Power\;=\;\bf{\dfrac{Workdone}{Time}}}}}$

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★ Solution :-

Given,

» Mass of the coal = m = 105 Kg

» Time taken by engine = t = 1 hr = 3600 seconds

» Height from which coal is lifted = h = 360 m

» Acceleration due to gravity = g = 10 m/sec²

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~ For the Workdone by engine by lifting the weight ::

We know that, Potential Energy of coal = Workdone by machine to lift it.

So firstly we will calculate the Potential Energy of the coal. This is given as,

$\\\;\sf{\rightarrow\;\;Potential\;Energy\;=\;\bf{m\:g\:h}}$

By applying values, we get

$\\\;\sf{\rightarrow\;\;Potential\;Energy\;of\;Coal\;=\;\bf{105\:\times\:360\:\times\:10}}$

$\\\;\bf{\rightarrow\;\;Potential\;Energy\;of\;Coal\;=\;\bf{\green{378000\;\:J}}}$

Here J shows the SI unit of Workdone that is Joules because here all the units are in SI so the resultant unit is also in SI.

Now we also know that, SI unit of Workdone is Joules. So,

$\\\;\bf{\rightarrow\;\;Workdone\;by\;Engine\;=\;\bf{\blue{378000\;\:J}}}$

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~ For the Power of the Engine ::

We know that,

$\\\;\sf{\rightarrow\;\;Power\;=\;\bf{\dfrac{Workdone}{Time}}}$

Now by applying value, we get

$\\\;\sf{\Longrightarrow\;\;Power\;=\;\bf{\dfrac{378000}{3600}}}$

$\\\;\sf{\Longrightarrow\;\;Power\;=\;\bf{\dfrac{3780}{36}}}$

$\\\;\bf{\Longrightarrow\;\;Power\;=\;\bf{\red{105\;\:Watts}}}$

$\\\;\underline{\boxed{\tt{Power\;of\;the\;Engine\;=\;\bf{\purple{105\;\;Watts}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;Force\;=\;mass\:\times\:Acceleration}$

$\\\;\sf{\leadsto\;\;Workdone\;=\;Force\:\times\:Distance}$

$\\\;\sf{\leadsto\;\;Kinetic\;Energy\;=\;\dfrac{1}{2}\:m\:v^{2}}$

$\\\;\sf{\leadsto\;\;Pressure\;=\;\dfrac{Force}{Area}}$

$\\\;\sf{\leadsto\;\;Momentum\;=\;Mass\:\times\:Velocity}$