Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take g constant with a value 10 m/s². The work done by the (i) gravitational force and the (ii) resistive force of air is

Question

The energy associated with electric field is E and with magnetic field is B for an electromagnetic wave in free space is given by( \(∈_0\)-permittivity of free space \(μ_0\)-permeability of free space)

Answer 1

To solve this problem, we need to determine the work done by the gravitational force and the resistive force of air on a raindrop falling from a certain height.

Given data:

Mass of the raindrop, \(m = 1 \, \text{g} = 0.001 \, \text{kg}\)
Height from which the drop falls, \(h = 1 \, \text{km} = 1000 \, \text{m}\)
Final speed of the raindrop, \(v = 50 \, \text{m/s}\)
Gravitational acceleration, \(g = 10 \, \text{m/s}^2\)

Let's break it down into parts:

Work Done by Gravitational Force:
- The work done by gravity is given by the formula: \(W_{\text{gravity}} = mgh\)
- Substituting the values, we get: \(W_{\text{gravity}} = 0.001 \times 10 \times 1000 = 10 \, \text{J}\)
- Thus, the work done by the gravitational force is 10 J.
Work Done by Resistive Force of Air:
- The work-energy principle states that the change in kinetic energy (ΔKE) of the object is equal to the net work done on the object.
- Change in kinetic energy, \(\Delta KE = \frac{1}{2}mv^2 - 0\) (initial kinetic energy is zero since it starts from rest)
- Substituting the values, \(\Delta KE = \frac{1}{2} \times 0.001 \times (50)^2 = \frac{1}{2} \times 0.001 \times 2500 = 1.25 \, \text{J}\)
- Total work done by all forces is the change in kinetic energy: \(W_{\text{gravity}} + W_{\text{resistive}} = \Delta KE\)
- So, \(10 + W_{\text{resistive}} = 1.25\)
- Solving for \(W_{\text{resistive}}\), we get: \(W_{\text{resistive}} = 1.25 - 10 = -8.75 \, \text{J}\)
- The negative sign indicates that the work done by the resistive force is against the direction of motion.