Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

A raindrop of mass \[1\,{\text{g}}\] falling from height of \[1\,{\text{km}}\] hits the ground with a speed of \[50\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\]. If the resistive force is proportional to the speed of the drop, then the work done by the resistive force is (Take \[g = 10\,{\text{m}} \cdot {{\text{s}}^{ - 2}}\])
A. \[10\,{\text{J}}\]
B. \[ - 10\,{\text{J}}\]
C. \[8.75\,{\text{J}}\]
D. \[ - 8.75\,{\text{J}}\]

seo-qna
Last updated date: 13th Jun 2024
Total views: 402.6k
Views today: 6.02k
Answer
VerifiedVerified
402.6k+ views
Hint: Use work-energy theorem for the raindrop. The expression for this theorem gives the relation between the change in kinetic energy of the raindrop and the total work done by all the forces on the raindrop

Formula used:
The expression for the kinetic energy of an object is
\[K = \dfrac{1}{2}m{v^2}\] …… (1)
Here, \[K\] is the kinetic energy of the object, \[m\] is the mass of the object and \[v\] is the velocity of the object.
The expression for the potential energy of an object is
\[U = mgh\] …… (2)
Here, \[U\] is the potential energy of the object, \[m\] is the mass of the object, \[g\] is the acceleration due to gravity and \[h\] is the height of the object from the ground.
The expression for the work-energy theorem is
\[W = \Delta K\] …… (3)
Here, \[W\] is the work done by an object and \[\Delta K\] is the change in the kinetic energy of the object

Complete step by step answer:
The raindrop has mass \[1\,{\text{g}}\] and it is falling from a height of \[1\,{\text{km}}\] hits the ground with a speed \[50\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\].
The initial speed of the raindrop at height \[1\,{\text{km}}\] is zero as it has zero kinetic energy at the maximum height.
Calculate the change in kinetic energy of the raindrop during its travel.
Rewrite equation (1) for the initial and final kinetic energies of the raindrop at maximum height and the ground.
\[{K_i} = \dfrac{1}{2}m{u^2}\]
Here, \[{K_i}\] is the initial kinetic energy of the raindrop, \[m\] is the mass of the raindrop and \[u\] is the initial speed of the raindrop at maximum height.
\[{K_f} = \dfrac{1}{2}m{v^2}\]
Here, \[{K_f}\] is the final kinetic energy of the raindrop and \[v\] is the final speed of the raindrop when it just touches the ground.
The change in the kinetic energy \[\Delta K\] of the raindrop is the subtraction of the initial and final kinetic energies of the raindrop.
\[\Delta K = {K_i} - {K_f}\]
Substitute \[\dfrac{1}{2}m{u^2}\] for \[{K_i}\] and \[\dfrac{1}{2}m{v^2}\] for \[{K_f}\] in the above equation.
\[\Delta K = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{v^2}\]
\[ \Rightarrow \Delta K = \dfrac{1}{2}m\left( {{u^2} - {v^2}} \right)\]
Substitute \[{\text{1}}\,{\text{g}}\] for \[m\], \[0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\] for \[u\] and \[50\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\] for \[v\] in the above equation.
\[ \Rightarrow \Delta K = \dfrac{1}{2}\left( {{\text{1}}\,{\text{g}}} \right)\left[ {{{\left( {0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)}^2} - {{\left( {50\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)}^2}} \right]\]
\[ \Rightarrow \Delta K = \dfrac{1}{2}\left( {{\text{1}} \times {\text{1}}{{\text{0}}^{ - 3}}\,{\text{kg}}} \right)\left[ { - {{\left( {50\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)}^2}} \right]\]
\[ \Rightarrow \Delta K = - 1.25\,{\text{J}}\]
Hence, the change in kinetic energy of the raindrop during its travel is \[ - 1.25\,{\text{J}}\].
The total work done on the raindrop is the sum of the work done by the gravitational force and the work done by the resistive force.
The work done by the gravitational force on the raindrop is equal to the potential energy of the raindrop at its maximum height.
Calculate the work done \[{W_g}\] by the gravitational force on the raindrop.
Substitute \[{W_g}\] for \[U\] in equation (2).
\[{W_g} = mgh\]
Substitute \[{\text{1}}\,{\text{g}}\] for \[m\], \[10\,{\text{m}} \cdot {{\text{s}}^{ - 2}}\] for \[g\] and \[1\,{\text{km}}\] for \[h\] in equation (2).
\[{W_g} = \left( {{\text{1}}\,{\text{g}}} \right)\left( {10\,{\text{m}} \cdot {{\text{s}}^{ - 2}}} \right)\left( {1\,{\text{km}}} \right)\]
\[ \Rightarrow {W_g} = \left( {{\text{1}} \times {\text{1}}{{\text{0}}^{ - 3}}\,{\text{kg}}} \right)\left( {10\,{\text{m}} \cdot {{\text{s}}^{ - 2}}} \right)\left( {1 \times {\text{1}}{{\text{0}}^3}\,{\text{m}}} \right)\]
\[ \Rightarrow {W_g} = 10\,{\text{J}}\]
Hence, the work done by the gravitational force on the raindrop is \[10\,{\text{J}}\].
Now, calculate the work done by the resistive force.
According to the work-energy theorem, the change in the kinetic energy of the raindrop is equal to the total work done on the raindrop by the resistive force and the gravitational force.
\[{W_r} + {W_g} = \Delta K\]
Here, \[{W_r}\] is the work done by the resistive force.
Rearrange the above equation for \[{W_r}\].
\[{W_r} = \Delta K - {W_g}\]
Substitute \[1.25\,{\text{J}}\] for \[\Delta K\] and \[10\,{\text{J}}\] for \[{W_g}\] in the above equation.
\[{W_r} = \left( {1.25\,{\text{J}}} \right) - \left( {10\,{\text{J}}} \right)\]
\[{W_r} = - 8.75\,{\text{J}}\]
Hence, the work done by the resistive force on the raindrop is \[ - 8.75\,{\text{J}}\].

So, the correct answer is “Option D”.

Note:
The negative sign of the work done by the resistive force on the raindrop indicates that the resistive force on the raindrop is in the opposite direction as that of the gravitational force on the raindrop.