Question

A raindrop of mass$\mathbf{3}\mathbf{.}\mathbf{35}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathit{k}\mathit{g}$ falls vertically at constant speed under the influence of gravity and air resistance. Model the drop as a particle. As it falls , $\mathbf{100}\mathit{m}$what is the work done on the raindrop

(a) by the gravitational force and

(b) by air resistance?

Short answer

(a) The work done by the gravitational force on the rain drop is $\mathbf{32}\mathbf{.}\mathbf{83}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{J}$.

(b) The work done bythe frictional force on the rain drop is .$\mathbf{-}\mathbf{32}\mathbf{.}\mathbf{83}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{J}$

Step-by-step solution

Given data

The mass of the rain drop is

$m=3.35\times {10}^{-5}\text{\hspace{0.17em}kg}$

Displacement of the drop is

$s=100\text{\hspace{0.17em}m}$

Work done and force due to gravity

Work done by a force$F$ causing displacement $s$ at an angle $\theta$ with the force is

$\mathit{W}\mathbf{=}\mathit{F}\mathit{s}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$ .....(I)

The force due to gravity is on a particle of mass $m$ is

$\mathit{F}\mathbf{=}\mathit{m}\mathit{g}$ .....(II)

Here $g$ is the acceleration due to gravity of value

$\mathit{g}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{8}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$

Determining the work done by gravity

From equation (II) the force applied by gravity on the drop is

$\begin{array}{c}F=3.35\times {10}^{-5}\text{\hspace{0.17em}kg}\times 9.8{\text{\hspace{0.17em}m/s}}^2\\ =32.83\times {10}^{-5}\cdot (1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^2\times \frac{1\text{\hspace{0.17em}N}}{1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^2})\\ =32.83\times {10}^{-5}\text{\hspace{0.17em}N}\end{array}$

The force due to gravity and the displacement of the drop are both directed vertically downward. Thus from equation (I) the work done by gravity is

$\begin{array}{c}W=32.83\times {10}^{-5}\text{\hspace{0.17em}N}\times 100\text{\hspace{0.17em}m}\times \text{cos0}\\ =32.83\times {10}^{-3}\cdot (1\text{\hspace{0.17em}N}\cdot \text{m}\times \frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}N}\cdot \text{m}})\\ =32.83\times {10}^{-3}\text{\hspace{0.17em}J}\end{array}$

The required work done is.$32.83\times {10}^{-3}\text{\hspace{0.17em}J}$

Determining the work done by friction

The rain drop is moving at a constant speed. Thus the net force on it is zero. Hence the magnitude of gravitational force and frictional force both have the same magnitude but oppositely directed. The work done by the frictional force is thus

$\begin{array}{c}W=32.83\times {10}^{-5}\text{\hspace{0.17em}N}\times 100\text{\hspace{0.17em}m}\times \text{cos180}°\\ =-32.83\times {10}^{-3}\cdot (1\text{\hspace{0.17em}N}\cdot \text{m}\times \frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}N}\cdot \text{m}})\\ =-32.83\times {10}^{-3}\text{\hspace{0.17em}J}\end{array}$

Thus the required work done is .$-32.83\times {10}^{-3}\text{\hspace{0.17em}J}$