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$^2$)
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$^2$)
(take $g = 10$ m/s$^2$)
Show Hint
Apply the Work-Energy Theorem: The total work done on an object is equal to the change in its kinetic energy.
Updated On: Apr 9, 2026
- $-8.75$
- $-8.35$
- $-9.55$
- $-9.98$
Show Solution
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The Correct Option is D
Solution and Explanation
The problem involves calculating the work done by a resistive force (like air resistance) on a falling object. We can use the Work-Energy Theorem, which states that the total work done by all forces acting on an object is equal to the change in its kinetic energy.
$$W_{\text{total}} = \Delta K$$
The forces acting on the raindrop are gravity and the resistive force. Therefore:
$$W_{\text{gravity}} + W_{\text{resistive}} = K_{\text{final}} - K_{\text{initial}}$$
Step 1: Identify given values in SI units.
Mass $m = 1$ g $= 10^{-3}$ kg
Initial velocity $u = 0$ m/s
Final velocity $v = 5$ m/s
Height $h = 1$ km $= 1000$ m
Acceleration due to gravity $g = 10$ m/s$^2$
Step 2: Calculate work done by gravity.
$$W_{\text{gravity}} = mgh = (10^{-3}) \times (10) \times (1000) = 10 \text{ J}$$
Step 3: Calculate the change in kinetic energy.
$$K_{\text{initial}} = \frac{1}{2} m u^2 = 0$$
$$K_{\text{final}} = \frac{1}{2} m v^2 = \frac{1}{2} \times 10^{-3} \times (5)^2 = 0.5 \times 0.001 \times 25 = 0.0125 \text{ J}$$
$$\Delta K = 0.0125 - 0 = 0.0125 \text{ J}$$
Step 4: Solve for $W_{\text{resistive}}$.
$$10 + W_{\text{resistive}} = 0.0125$$
$$W_{\text{resistive}} = 0.0125 - 10 = -9.9875 \text{ J}$$
Rounding to two decimal places, we get $-9.98$ J.
$$W_{\text{total}} = \Delta K$$
The forces acting on the raindrop are gravity and the resistive force. Therefore:
$$W_{\text{gravity}} + W_{\text{resistive}} = K_{\text{final}} - K_{\text{initial}}$$
Step 1: Identify given values in SI units.
Mass $m = 1$ g $= 10^{-3}$ kg
Initial velocity $u = 0$ m/s
Final velocity $v = 5$ m/s
Height $h = 1$ km $= 1000$ m
Acceleration due to gravity $g = 10$ m/s$^2$
Step 2: Calculate work done by gravity.
$$W_{\text{gravity}} = mgh = (10^{-3}) \times (10) \times (1000) = 10 \text{ J}$$
Step 3: Calculate the change in kinetic energy.
$$K_{\text{initial}} = \frac{1}{2} m u^2 = 0$$
$$K_{\text{final}} = \frac{1}{2} m v^2 = \frac{1}{2} \times 10^{-3} \times (5)^2 = 0.5 \times 0.001 \times 25 = 0.0125 \text{ J}$$
$$\Delta K = 0.0125 - 0 = 0.0125 \text{ J}$$
Step 4: Solve for $W_{\text{resistive}}$.
$$10 + W_{\text{resistive}} = 0.0125$$
$$W_{\text{resistive}} = 0.0125 - 10 = -9.9875 \text{ J}$$
Rounding to two decimal places, we get $-9.98$ J.
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