Question

The rain drop of mass $1$ g, starts with zero velocity from a height of $1$ km. It hits the ground with a speed of $5$ m/s. The work done by the unknown resistive force is —————— J.
(take $g = 10$ m/s$^2$)

• A. $-8.75$
• B. $-8.35$
• C. $-9.55$
• D. $-9.98$

Hint:

Apply the Work-Energy Theorem: The total work done on an object is equal to the change in its kinetic energy.

Answer 1

The Correct Option is D

We apply the principle of conservation of energy including non-conservative forces. The energy balance is given by:
Initial Potential Energy + Initial Kinetic Energy + Work done by non-conservative force = Final Potential Energy + Final Kinetic Energy
Let the ground be the reference for potential energy ($PE = 0$ at ground level).
Step 1: Initial state (at height $h$):
$PE_i = mgh = (0.001 \text{ kg}) \times (10 \text{ m/s}^2) \times (1000 \text{ m}) = 10 \text{ J}$
$KE_i = \frac{1}{2} m u^2 = 0$ (since it starts from rest)
Step 2: Final state (at ground level):
$PE_f = 0$
$KE_f = \frac{1}{2} m v^2 = \frac{1}{2} \times (0.001 \text{ kg}) \times (5 \text{ m/s})^2 = 0.5 \times 0.001 \times 25 = 0.0125 \text{ J}$
Step 3: Energy equation:
$$10 + 0 + W_{\text{resistive}} = 0 + 0.0125$$
$$W_{\text{resistive}} = 0.0125 - 10 = -9.9875 \text{ J}$$
The negative sign indicates that the resistive force acts opposite to the direction of motion, thus removing energy from the system. Comparing with the options, the value is approximately $-9.98$ J.