Question

It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of drop but is otherwise undetermined. Consider a drop of mass $1.00g$ falling from a height $1.00km$. It hits the ground with a speed of $50.0m{s^{ - 1}}$. What is the work done by the unknown resistive force?

Answer 1

As the raindrop is falling under the influence of gravity the acceleration of the drop is equal to acceleration due to gravity, height and mass of the raindrop is given so we can find the work done by the gravitational force. At the same time we know the speed of the raindrop now we can calculate the kinetic energy of the raindrop. After all this then we can calculate the work done by the resistive force by applying work energy theorem.

Complete answer:
As per given problem a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of drop but is otherwise undetermined. Consider a drop of mass $1.00g$ falling from a height $1.00km$. It hits the ground with a speed of $50.0m{s^{ - 1}}$.
As the raindrop is hitting the ground with a speed of $50.0m{s^{ - 1}}$, the mass of the drop is $1.00g$ and the initial velocity of the body is zero.
Now we can calculate the change in kinetic energy of the drop.
Change in kinetic energy = final kinetic energy – initial kinetic energy
The initial kinetic energy of the rain drop is zero as the initial velocity is zero.
Now the change in kinetic energy will become,
Change in kinetic energy = final kinetic energy
$\Delta K.E = \dfrac{1}{2}m{v^2}$
We know,
$v = 50.0m{s^{ - 1}}$
$m = 1.00g = {10^{ - 3}}kg$
SI unit of mass is kg.
Putting the given values in the change in kinetic energy formula we will get,
$\Delta K.E = \dfrac{1}{2} \times {10^{ - 3}}kg \times {\left( {50.0m{s^{ - 1}}} \right)^2}$
$\Rightarrow \Delta K.E = \dfrac{1}{2} \times {10^{ - 3}}kg \times 2500.0{m^2}{s^{ - 2}}$
Cancelling the terms we will get,
$\Delta K.E = {10^{ - 3}}kg \times 1250.0{m^2}{s^{ - 2}}$
$\Rightarrow \Delta K.E = 1.25J \ldots \ldots \left( 1 \right)$
As the raindrop falls under the influence of gravitational force then the acceleration due to gravity will be g which is equal to $10m{s^{ - 2}}$.
The work done by the gravitational force is,
$Wg = mgh$
Where,
Mass of the raindrop = $m = 1.00g = {10^{ - 3}}kg$
Acceleration due gravity = $g = 10m{s^{ - 2}}$
Height from which the raindrop is falling under the influence of gravitational force = $1.00km = {10^3}m$
SI unit of height is meter.
Putting the given value in the work done formula we will get,
$Wg = {10^{ - 3}}kg \times 10m{s^{ - 2}} \times {10^3}m$
On further solving we will get,
$Wg = 10.0J \ldots \ldots \left( 2 \right)$
According the work energy theorem,
Change in kinetic energy = work done due to external force
Mathematically,
$\Delta K.E = Wr + Wg \ldots \ldots \left( 3 \right)$
Where,
$Wr$ = Work Done due to resistive force
Putting equation $\left( 1 \right)$ and $\left( 2 \right)$ in equation $\left( 3 \right)$ we will get,
$1.25J = Wr + 10J$
Rearranging the above equation we will get,
$\Rightarrow 1.25J - 10.0J = Wr$
$\therefore Wr = - 8.75J$ is negative.

Note:
Before putting the given value first, change all values in its SI term so as to get the kinetic energy and work done in its basic unit. Remember that the negative work done due to resistive force denotes that the work done is done against the work done due to gravitational force. Also note that the initial velocity of the raindrop is taken as zero as the raindrop is initially at rest.