- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
(a) Explain by an example what is meant by potential energy. Write down the expression for gravitational potential energy of a body of mass m placed at a height h above the surface of the earth.
(b) What is the difference between potential energy and kinetic energy?
(c) A ball of mass $0.5\ kg$ slows down from a speed of $5\ m/s$ to that of $3\ m/s$. Calculate the change in kinetic energy of the ball. State your answer giving proper units.
(a). Potential Energy: The energy of a body due to its position is known as the gravitational potential energy of the body.
If a body of mass $m$ is placed at a height $h$ above the surface of the earth, then its potential energy $P=mgh$
Here, $g\rightarrow$gravitational acceleration
(b). Difference between potential energy and kinetic energy:
Potential Energy | Kinetic Energy |
1. A body possesses potential energy due its virtue of positions. | 1. A body possesses kinetic energy due to its motion. |
2. It depends on the height of the body from the surface of the ground. | 2. It depends on the velocity of the of the moving body. |
3. A stone placed on the roof is a good example of possessing potential energy. | 3. A car moving on the road is a good example of bdy possessing kinetic energy. |
(c). Mass $m=0.5\ kg$
Initial speed $v_1=5\ m/s$
Final speed $v_2=3\ m/s$
Change in kinetic energy $=K_1-K_2$
$=\frac{1}{2}mv_1^2-\frac{1}{2}mv_2^2$
$=\frac{1}{2}m(v_1^2-v_2^2)$
$=\frac{1}{2}\times0.5\times(5^2-3^2)$
$=\frac{1}{2}\times0.5\times(25-9)$
$=\frac{1}{2}\times0.5\times16$
$=4.0\ J$
Thus, there is a change in kinetic energy of $4.0\ J$