Question

A rain drop of mass \(10g\) falls from a height of \(50m\) from rest. If the loss of energy due to air resistance is \(3J\), then the velocity of the drop on striking the ground is \((g = 10ms^{-2})\)

• A. \(10 m s^{-1}\)
• B. \(5 ms^{-1}\)
• C. \(30 ms^{-1}\)
• D. \(40 ms^{-1}\)
• E. \(20 ms^{-1}\)

Hint:

Always convert mass to kg and ensure consistent units when applying the work-energy theorem.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
This problem involves the principle of conservation of energy, taking into account the work done by a non-conservative force (air resistance). The initial potential energy of the raindrop is converted into kinetic energy and heat due to air resistance.
Step 2: Key Formula or Approach:
The work-energy theorem states that the change in mechanical energy is equal to the work done by non-conservative forces. \[ E_{initial} = E_{final} + E_{lost} \] Here, the initial energy is purely gravitational potential energy (PE), the final energy is purely kinetic energy (KE), and the energy lost is due to air resistance. \[ \text{PE}_{initial} = \text{KE}_{final} + E_{lost} \] \[ mgh = \frac{1}{2}mv^2 + E_{lost} \] Step 3: Detailed Explanation:
First, we list the given values and convert them to SI units.
Mass, \(m = 10 \text{ g} = 10 \times 10^{-3} \text{ kg} = 0.01 \text{ kg}\)
Height, \(h = 50 \text{ m}\)
Acceleration due to gravity, \(g = 10 \text{ m/s}^2\)
Energy lost due to air resistance, \(E_{lost} = 3 \text{ J}\)
Next, we calculate the initial potential energy (PE) of the raindrop. \[ \text{PE}_{initial} = mgh = (0.01 \text{ kg})(10 \text{ m/s}^2)(50 \text{ m}) = 5 \text{ J} \] Now, we use the energy conservation equation to find the final kinetic energy (KE) just before it hits the ground. \[ \text{KE}_{final} = \text{PE}_{initial} - E_{lost} \] \[ \text{KE}_{final} = 5 \text{ J} - 3 \text{ J} = 2 \text{ J} \] Finally, we use the formula for kinetic energy to find the final velocity (v). \[ \text{KE}_{final} = \frac{1}{2}mv^2 \] \[ 2 = \frac{1}{2}(0.01)v^2 \] \[ 4 = 0.01 \times v^2 \] \[ v^2 = \frac{4}{0.01} = 400 \] \[ v = \sqrt{400} = 20 \text{ m/s} \] Step 4: Final Answer:
The velocity of the drop on striking the ground is 20 ms\(^{-1}\). This corresponds to option (E).