A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s¯¹?
Given: The radius of the rain drop is 2 mm , the initial height of the rain drop from the ground is 500 m and speed of the drop on reaching the ground is 10 ms -1 .
The volume of the rain drop can be given as,
V = 4 3 π r 3
where, ris the radius of the drop.
By substituting the given values in the above expression, we get
V = 4 3 π ( 2 × 10 − 3 ) 3 m 3
We know that the density of water is ρ = 10 3 kg / m 3 .
The mass of the rain drop is given as,
m = ρ V
By substituting the values of volume and density in the above expression, we get
m = 4 3 π ( 2 × 10 − 3 ) 3 × 10 3 kg
Gravitational force F on the rain drop is given as,
F = m g
Where, g is the acceleration due to gravity ( 9.8 m / s ) .
By substituting the values of mass of the rain drop and acceleration due to gravity in the above expression, we get
F = 4 3 π ( 2 × 10 − 3 ) 3 × 10 3 × 9.8 N
The work done by the gravitational force on the rain drop in the first half of the journey is given as,
W a = F ⋅ s
By substituting the values of gravitational force anddistance covered by the rain drop during first half of its journey in the above expression, we get
W a = ( 4 3 π ( 2 × 10 − 3 ) 3 × 10 3 × 9.8 N ) ( 250 m ) = 0 .082 J
Since, the distance covered by the rain drop in first half of the journey and the second half of journey is the same. Therefore, the work done by the gravitational force on the drop in the first half of the journey will be equal to that in thesecond half of the journey.
The work done by the gravitational force in the second half of the journey is given as,
W b = 0.082 J
The total energy of the rain drop is equal to the potential energy of the rain drop at the height of 500 m .
E = m g h
Where, E is the total energy and h is the height.
By substituting the given values in the above equation, we get
E = ( 4 3 π ( 2 × 10 − 3 ) 3 × 10 3 kg ) ( 9.8 m/s 2 ) ( 500 m ) = 0.164 J
When the rain drop hits the ground, the kinetic energy K . E . of the drop is given as,
K . E . = 1 2 m v 2
By substituting the value of mass of the rain drop and the velocity with which the rain drop hits the ground in the above expression, we get
K . E . = 1 2 × ( 4 3 π ( 2 × 10 − 3 ) 3 × 10 3 kg ) × ( 10 m/s ) 2 = 1.675 × 10 − 3 J
The work done by the resistive force will be equal to the difference inthe initial potential energy of the rain drop to the final kinetic energy of the rain drop. Therefore,
W R = E − K E = ( 0.164 J ) − ( 1.675 × 10 − 3 J ) = − 0.162 J
Thus, the work done by the resistive force in the entire journey is − 0.162 J .
Given: The radius of the rain drop is
The volume of the rain drop can be given as,
where, ris the radius of the drop.
By substituting the given values in the above expression, we get
We know that the density of water is
The mass of the rain drop is given as,
By substituting the values of volume and density in the above expression, we get
Gravitational force
Where,
By substituting the values of mass of the rain drop and acceleration due to gravity in the above expression, we get
The work done by the gravitational force on the rain drop in the first half of the journey is given as,
By substituting the values of gravitational force anddistance covered by the rain drop during first half of its journey in the above expression, we get
Since, the distance covered by the rain drop in first half of the journey and the second half of journey is the same. Therefore, the work done by the gravitational force on the drop in the first half of the journey will be equal to that in thesecond half of the journey.
The work done by the gravitational force in the second half of the journey is given as,
The total energy of the rain drop is equal to the potential energy of the rain drop at the height of
Where,
By substituting the given values in the above equation, we get
When the rain drop hits the ground, the kinetic energy
By substituting the value of mass of the rain drop and the velocity with which the rain drop hits the ground in the above expression, we get
The work done by the resistive force will be equal to the difference inthe initial potential energy of the rain drop to the final kinetic energy of the rain drop. Therefore,
Thus, the work done by the resistive force in the entire journey is
