30. A raindrop of mass 2 g is falling from a height Jobne of 1.00 km. It is the ground with a speed of 60.0 ms-1. Find the work done:

(A) by the gravitational force

(B) by the unknown resistive force

molga

(g = 10 m/s²)

To solve this problem, we need to calculate the work done by the gravitational force and the work done by the unknown resistive force on the falling raindrop.

### Given:
- Mass of the raindrop: \( m = 2 \, \text{g} = 2 \times 10^{-3} \, \text{kg} \)
- Initial height: \( h = 1.00 \, \text{km} = 1000 \, \text{m} \)
- Final speed: \( v = 60.0 \, \text{m/s} \)
- Gravitational acceleration: \( g = 10 \, \text{m/s}^2 \)

### (A) Work done by the gravitational force

The work done by gravity is calculated using the formula:

\[
W_{\text{gravity}} = mgh
\]

Substitute the known values:

\[
W_{\text{gravity}} = (2 \times 10^{-3} \, \text{kg}) \times (10 \, \text{m/s}^2) \times (1000 \, \text{m})
\]

\[
W_{\text{gravity}} = 20 \, \text{J}
\]

So, the work done by the gravitational force is **20 joules**.

### (B) Work done by the unknown resistive force

The work done by the resistive force can be determined using the work-energy principle, which states that the net work done on an object is equal to the change in its kinetic energy.

The change in kinetic energy \( \Delta K \) is given by:

\[
\Delta K = \frac{1}{2} m v^2 - \frac{1}{2} m u^2
\]

where:
- \( v = 60.0 \, \text{m/s} \) is the final speed,
- \( u = 0 \, \text{m/s} \) is the initial speed (since the raindrop starts from rest).

Substitute the values:

\[
\Delta K = \frac{1}{2} (2 \times 10^{-3} \, \text{kg}) \times (60.0 \, \text{m/s})^2
\]

\[
\Delta K = \frac{1}{2} (2 \times 10^{-3}) \times 3600
\]

\[
\Delta K = 3.6 \, \text{J}
\]

Now, the total work done on the raindrop is the sum of the work done by gravity and the work done by the resistive force. Since the net work done is equal to the change in kinetic energy, we can write:

\[
W_{\text{net}} = W_{\text{gravity}} + W_{\text{resistive}}
\]

Since \( W_{\text{net}} = 3.6 \, \text{J} \), we can rearrange the equation to solve for the work done by the resistive force:

\[
W_{\text{resistive}} = \Delta K - W_{\text{gravity}}
\]

\[
W_{\text{resistive}} = 3.6 \, \text{J} - 20 \, \text{J}
\]

\[
W_{\text{resistive}} = -16.4 \, \text{J}
\]

The negative sign indicates that the resistive force opposes the motion of the raindrop, doing negative work.

### Final Answer:
- (A) The work done by the gravitational force is \( 20 \, \text{J} \).
- (B) The work done by the unknown resistive force is \( -16.4 \, \text{J} \).

Would you like more details or have any other questions?

### Related Questions:
1. How does the resistive force affect the velocity of the falling object?
2. What would the final speed of the raindrop be without any resistive force?
3. How would the work done by gravity change if the height were doubled?
4. Can you explain the work-energy theorem in detail?
5. What is the relationship between gravitational potential energy and kinetic energy?

### Tip:
The work-energy theorem is a powerful tool for analyzing the motion of objects when forces, including resistive ones, act on them. Always check if the system loses or gains mechanical energy.

A raindrop of mass 2 g is falling from a height of 1.00 km. It hits the ground with a speed of 60.0 m/s. Find the work done: (A) by the gravitational force (B) by the unknown resistive force (g = 10 m/s²)

Learn how to calculate the work done by gravitational and resistive forces on a falling raindrop. The problem explores key physics concepts like the work-energy principle, kinetic energy, and gravitational potential energy. Suitable for high school students (Grades 10-12).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation