A raindrop 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 ms-1?

In solving this problem, we'll consider the work done by the gravitational force and the resistive force separately.

1. Work done by the gravitational force: Work done by gravity is given by the formula: W=F⋅d⋅cos⁡(θ)W=F⋅d⋅cos(θ), where FF is the force, dd is the displacement, and θθ is the angle between the force and displacement vectors.

   In the first half of the journey, the raindrop is accelerating due to gravity. The work done by gravity during this period can be calculated using the formula W1=mgh1W1=mgh1, where mm is the mass of the raindrop, gg is the acceleration due to gravity, and h1h1 is the initial height.

   In the second half of the journey, the raindrop is moving with uniform speed, so the gravitational force does no work during this period.

2. Work done by the resistive force: The resistive force (viscous drag) opposes the motion of the raindrop. When the raindrop reaches its terminal velocity, the resistive force equals the gravitational force, and thus no net force acts on the raindrop. So, in the second half of the journey, the work done by the resistive force is zero.

   However, during the first half of the journey, the resistive force acts in the opposite direction to the displacement. The work done by the resistive force can be calculated using the formula W2=−FdW2=−Fd, where FF is the resistive force and dd is the displacement.

Now, let's calculate each part step by step.

1. Work done by gravity in the first half: Given:

   • Initial height, h1=500h1=500 m
   • Radius of raindrop, r=2r=2 mm
   • Mass of the raindrop, m=43πr3ρm=34πr3ρ, where ρρ is the density of water (about 1000 kg/m31000kg/m3)
   • Acceleration due to gravity, g=9.8 m/s2g=9.8m/s2

   First, we calculate the mass of the raindrop: m=43π(0.002 m)3×1000 kg/m3m=34π(0.002m)3×1000kg/m3 m≈1.67×10−5 kgm≈1.67×10−5kg

   Then, we calculate the work done by gravity: W1=mgh1W1=mgh1 W1=(1.67×10−5 kg)(9.8 m/s2)(250 m)W1=(1.67×10−5kg)(9.8m/s2)(250m) W1≈0.041 JW1≈0.041J

2. Work done by the resistive force: Since the raindrop reaches terminal velocity in the first half of the journey, the resistive force is equal to the gravitational force. So, in the first half, the work done by the resistive force is equal in magnitude but opposite in direction to the work done by gravity, i.e., W2=−W1W2=−W1.

3. Work done by the resistive force in the entire journey: In the second half of the journey, the resistive force does no work as the raindrop moves with constant velocity. So, the total work done by the resistive force is just the negative of the work done by gravity in the first half of the journey, i.e., Wresistive=−W1Wresistive=−W1.

Therefore, the work done by the resistive force in the entire journey is approximately −0.041 J−0.041J, and the work done by gravity in the first half of the journey is approximately 0.041 J0.041J.