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In our daily lives, we often push, pull, lift, or move objects. These actions involve work, which is the effect of a force causing movement. When you lift a book, you use energy - the ability to do work. And when you do work quickly, you use power, which tells us how fast work is done.
Understanding work, energy, and power is important not only in science but also in solving problems in competitive exams like the BPSC Judiciary exam. These concepts help us explain how things move, how machines function, and how energy changes from one form to another.
Work is done when a force causes an object to move in the direction of the force. If you push a box and it moves, you have done work on the box.
The amount of work done depends on three things:
The formula to calculate work done is:
Here, cos θ means we only consider the part of the force that acts in the direction of the movement.
Energy is the capacity or ability to do work. Without energy, no work can be done. Energy exists in many forms, but two important types related to motion are kinetic energy and potential energy.
Kinetic energy is the energy an object has because it is moving. The faster it moves or the heavier it is, the more kinetic energy it has.
The formula for kinetic energy is:
Potential energy is stored energy an object has because of its position or height. For example, a book held up on a shelf has potential energy because it can fall down.
The formula for potential energy is:
As the pendulum swings, potential energy converts to kinetic energy and back, showing energy transformation.
Power tells us how fast work is done. If two people do the same amount of work, the one who does it faster has more power.
The formula for power is:
| Unit | Symbol | Equivalent |
|---|---|---|
| Watt | W | 1 Joule/second |
| Kilowatt | kW | 1000 Watts |
| Horsepower | hp | 746 Watts |
Understanding work, energy, and power helps us in many real-life situations:
Step 1: Identify the given values:
Step 2: Use the work formula:
\[ W = F \times d \times \cos\theta \]
Step 3: Calculate \( \cos 60^\circ = 0.5 \)
Step 4: Substitute values:
\[ W = 10 \times 5 \times 0.5 = 25 \, \text{J} \]
Answer: The work done is 25 Joules.
Step 1: Given:
Step 2: Use kinetic energy formula:
\[ KE = \frac{1}{2} m v^{2} \]
Step 3: Calculate \( v^{2} = 3^{2} = 9 \)
Step 4: Substitute values:
\[ KE = \frac{1}{2} \times 2 \times 9 = 9 \, \text{J} \]
Answer: The kinetic energy is 9 Joules.
Step 1: Given:
Step 2: Use potential energy formula:
\[ PE = m g h \]
Step 3: Substitute values:
\[ PE = 5 \times 9.8 \times 10 = 490 \, \text{J} \]
Answer: The potential energy is 490 Joules.
Step 1: Given:
Step 2: Use power formula:
\[ P = \frac{W}{t} \]
Step 3: Substitute values:
\[ P = \frac{500}{10} = 50 \, \text{W} \]
Answer: The power developed is 50 Watts.
Step 1: Given:
Step 2: According to the work-energy theorem, work done equals change in kinetic energy:
\[ W = KE_{final} - KE_{initial} \]
Since initial kinetic energy is zero,
\[ W = \frac{1}{2} m v^{2} \]
Step 3: Rearrange to find velocity \( v \):
\[ v = \sqrt{\frac{2W}{m}} \]
Step 4: Substitute values:
\[ v = \sqrt{\frac{2 \times 200}{20}} = \sqrt{20} = 4.47 \, \text{m/s} \]
Answer: The velocity of the object is approximately 4.47 m/s.
When to use: When given values in centimeters, grams, or minutes.
When to use: When force acts vertically but object moves horizontally or vice versa.
When to use: For pendulum or roller coaster type questions.
When to use: For power problems involving longer time units.
When to use: When force is not along the displacement direction.
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