Question

A rectangular trough, 2.0 m long, 0.70 m wide, and 0.40 m deep, is completely full of water. One end of the trough has a small drain plug right at the bottom edge.

When you pull the plug, at what speed does water emerge from the hole?

Answer 1

Concepts and reason

The concept used to solve the problem is law of conservation of energy. The potential energy due to the trough is conserved as the kinetic energy of the water column.

Initially, derive the expression for the speed of the water by using the conservation of energy. Later, substitute the values in the expression for the speed of the water to calculate the speed of the water.

Fundamentals

The potential energy of the water in the trough is expressed as:

$PE = mgh$

Here, m is the mass of the water, g is the acceleration due to gravity and h is the height of water column.

The kinetic energy of the water as it comes out with a certain speed is expressed as:

$KE = \frac{1}{2}m{v^2}$

Here, m is the mass of the water and v is the speed with which water comes out.

According to law of conservation of energy, the energy can neither be created nor destroyed. Energy can be changed from one form to other only.

The potential energy of the water is conserved as the kinetic energy of the water. The expression is given as:

$\begin{array}{c}\\PE = KE\\\\mgh = \frac{1}{2}m{v^2}\\\end{array}$

The speed of the water coming out is expressed as:

$\begin{array}{c}\\\frac{1}{2}m{v^2} = mgh\\\\v = \sqrt {2gh} \\\end{array}$

The speed of the water from the drain is calculated as:

$v = \sqrt {2gh}$

Substitute 0.4m for h, $9.8\;{\rm{m/}}{{\rm{s}}^2}$ for g and find the speed of the water.

$\begin{array}{c}\\v = \sqrt {2gh} \\\\ = \sqrt {2\left( {{\rm{9}}{\rm{.8}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {0.4\;{\rm{m}}} \right)} \\\\ = 2.8\;{\rm{m/s}}\\\end{array}$

The speed of the water from the drain is $2.8\;{\rm{m/s}}$

Ans:

The speed of the water from the drain is $2.8\;{\rm{m/s}}$