
An engine of $4.9kW$ is used to pump water from a $50m$ deep well. Calculate the amount of water drawn from the well if the pump is kept on for $1hr$.
Answer
123.3k+ views
Hint: The mass of water in kilograms is equal to the quantity of water in liters. The work done in pumping the water of the well will give you the amount of water drawn out of the well.
Complete step by step solution:
Power:
The work done by a certain force with respect to time is called power. A more appropriate definition is that the work done per unit time. Mathematically, it can be written as;
$P = \dfrac{W}{t}$
Here P is the power dissipated and W is the work done in time t.
So, we are given the power of the machine,
$P = 4.9kW$.
$ \Rightarrow P = 4900W$ (Given)
Also work is defined as the force done in moving the body by a certain distance.
Thus, mathematically, $W = F.x$
Now, we all know that the force here is mg and the distance is h,
So $P = \dfrac{{mgh}}{t}$
$ \Rightarrow P = \dfrac{{m(9.8)(50)}}{{3600}}$
$ \Rightarrow 4900 = \dfrac{{m(490)}}{{3600}}$
$ \Rightarrow m = \dfrac{{4900 \times 3600}}{{490}}$
$ \Rightarrow m = 36000kg$
Now let us convert the units;
$1L = 1000gram$
$ \Rightarrow 1L = 1kg$
$ \Rightarrow 1kL = 1000kg$
Thus, we get,
$V = 36kL$
Therefore the amount of water is 36 kiloliters.
Note:Unit conversion is something you must be very careful while doing.
The force drawing the water upwards is equal to the gravitational pulling force
This is because the net acceleration of the system is zero.
Complete step by step solution:
Power:
The work done by a certain force with respect to time is called power. A more appropriate definition is that the work done per unit time. Mathematically, it can be written as;
$P = \dfrac{W}{t}$
Here P is the power dissipated and W is the work done in time t.
So, we are given the power of the machine,
$P = 4.9kW$.
$ \Rightarrow P = 4900W$ (Given)
Also work is defined as the force done in moving the body by a certain distance.
Thus, mathematically, $W = F.x$
Now, we all know that the force here is mg and the distance is h,
So $P = \dfrac{{mgh}}{t}$
$ \Rightarrow P = \dfrac{{m(9.8)(50)}}{{3600}}$
$ \Rightarrow 4900 = \dfrac{{m(490)}}{{3600}}$
$ \Rightarrow m = \dfrac{{4900 \times 3600}}{{490}}$
$ \Rightarrow m = 36000kg$
Now let us convert the units;
$1L = 1000gram$
$ \Rightarrow 1L = 1kg$
$ \Rightarrow 1kL = 1000kg$
Thus, we get,
$V = 36kL$
Therefore the amount of water is 36 kiloliters.
Note:Unit conversion is something you must be very careful while doing.
The force drawing the water upwards is equal to the gravitational pulling force
This is because the net acceleration of the system is zero.
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