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Question: An engine can pump \(30000L\) water to a vertical height of \(45m\) in \(10\min \). Calculate the wo...

An engine can pump 30000L30000L water to a vertical height of 45m45m in 10min10\min . Calculate the work done by the machine and its power. (Given the density of water 103kg.m3{10^3}kg.{m^{ - 3}})

Explanation

Solution

First, we need to understand how to determine the work done by a machine and the relation between a machine's power and work done by a machine. The displacement done by a force measures the work done in a physical system. The power of a machine is given as the work done in a unit of time.

Complete step by step solution:
In physical science, work is referred to as the energy transferred from one object to another by applying a force. Work is represented as the product of applied force and displacement in the direction of that force. Work is a scalar quantity, although it is a product of two vector quantities.
In the simplest form, it is written as-
W=F.sW = F{{. }}s
where,
FF is the applied force
ss is the displacement along the direction of the applied force
Given,
The volume of water,
V=30000L=30000×103m3V = 30000L = 30000 \times {10^{ - 3}}{m^3} [1L=103m3]\left[ {\because 1L = {{10}^{ - 3}}{m^3}} \right]
The density of water, ρ=103kg.m3\rho = {10^3}kg.{m^{ - 3}}
Therefore, the mass of the given water-
M=ρ×VM = \rho \times V
M=30000×103×103kg\Rightarrow M = 30000 \times {10^{ - 3}} \times {10^3}kg
M=30000kg\Rightarrow M = 30000kg
We know work done by a machine is given by-
W=M.g.hW = M.g.h …………….(1)(1)
where,
gg is the gravitational acceleration
hh is the displacement done by the force
Therefore, we put g=9.8ms2g = 9.8m{s^{ - 2}} and h=45mh = 45m in the equation (1)(1)-
W=(30000×9.8×45)J\Rightarrow W = \left( {30000 \times 9.8 \times 45} \right)J
W=13.23×106J\Rightarrow W = 13.23 \times {10^6}J
Therefore, the total work done by the machine is 13.23×106J13.23 \times {10^6}J.
Given, the machine can pump the given amount of water in 10min10\min .
Therefore, the time,
t=10×60sec\Rightarrow t = 10 \times 60\sec
t=600sec\Rightarrow t = 600\sec
Now the power of a machine is given by-
P=WtP = \dfrac{W}{t}
We put W=13.23×106JW = 13.23 \times {10^6}J and t=600sect = 600\sec in the above equation and get-
P=13.23×106600W\Rightarrow P = \dfrac{{13.23 \times {{10}^6}}}{{600}}W
P=22050W\Rightarrow P = 22050W

Therefore, the power of the given machine is 22050W22050W.

Note: The most vital quantity related to a machine in a physical system is its mechanical efficiency. The efficiency of a machine can be given by the ratio of power output and power input. Mechanical efficiency is a dimensionless quantity that measures the effectiveness of the machine.