
A man is drawing water from a well with a bucket which leaks uniformly. The bucket when full weighs $20kg$ and when it arrives at the top only half the water remains. The depth of the water is 20metre. If $g=10m{{s}^{-2}}$ the work is done in $kJ$ will be:
Answer
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Hint: The work done by any object is equivalent to the product of the force and the displacement done by the body. When the work is done against gravity a decrease in the gravitational force is experienced by the body. And a downward force that is acting in the body $\left( mg \right)$ decreases as the bass of the object is reducing.
As per the given data,
The initial weight of the water is $20kg$
The final weight of the water is $10kg$
The depth of the water is $20m$
Complete step by step solution:
As per the situation mentioned in the question, water drowns from a leaking bucket. Initially, it has a weight of $20kg$ and over a distance of $20m$ the weight of the water reduces to $10kg$. So the rate of decrease of the mass per unit length will be,
$\begin{align}
& \dfrac{\left( 20-10 \right)kg}{20m} \\
& \Rightarrow 0.5kg\ per\ m \\
\end{align}$
The force acting on the body is given as the product of mass and acceleration of the body. When the body is doing some work concerning the ground its acceleration is the same as the acceleration due to gravity.
Mathematically,
$F=mg$
So the equation for the force of water whose mass is decreasing with the distance will be,
$dF=(200-0.5x)g$
Where,
$x$ Is the depth of the water covered by the bucket.
The work done by any object is given as the product of the force and the displacement
$W=F\times x$
For the work done over the displacement of $20m$ will be,
$\begin{align}
& W=\int_{0}^{20}{\left( 20-0.5x \right)g}dx \\
& \Rightarrow W=\left[ \left( 20-\dfrac{{{x}^{2}}}{4} \right)g \right]_{0}^{20} \\
& \therefore W=300J \\
\end{align}$
So, the required answer to the question is $300J$.
Note:
Displacement is the shortest path covered by an object. If the object reaches the origin point from where it starts to move the displacement is considered to be zero. If there are zero displacements the work done by the body will also be zero. The best example of such a case is a mam moving in a circular path.
As per the given data,
The initial weight of the water is $20kg$
The final weight of the water is $10kg$
The depth of the water is $20m$
Complete step by step solution:
As per the situation mentioned in the question, water drowns from a leaking bucket. Initially, it has a weight of $20kg$ and over a distance of $20m$ the weight of the water reduces to $10kg$. So the rate of decrease of the mass per unit length will be,
$\begin{align}
& \dfrac{\left( 20-10 \right)kg}{20m} \\
& \Rightarrow 0.5kg\ per\ m \\
\end{align}$
The force acting on the body is given as the product of mass and acceleration of the body. When the body is doing some work concerning the ground its acceleration is the same as the acceleration due to gravity.
Mathematically,
$F=mg$
So the equation for the force of water whose mass is decreasing with the distance will be,
$dF=(200-0.5x)g$
Where,
$x$ Is the depth of the water covered by the bucket.
The work done by any object is given as the product of the force and the displacement
$W=F\times x$
For the work done over the displacement of $20m$ will be,
$\begin{align}
& W=\int_{0}^{20}{\left( 20-0.5x \right)g}dx \\
& \Rightarrow W=\left[ \left( 20-\dfrac{{{x}^{2}}}{4} \right)g \right]_{0}^{20} \\
& \therefore W=300J \\
\end{align}$
So, the required answer to the question is $300J$.
Note:
Displacement is the shortest path covered by an object. If the object reaches the origin point from where it starts to move the displacement is considered to be zero. If there are zero displacements the work done by the body will also be zero. The best example of such a case is a mam moving in a circular path.
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