Work, Power, and Energy JEE Mains Questions
As a JEE Mains aspirant, you understand the importance of mastering important topics like work, power, and energy. These concepts not only form the foundation of physics but are also crucial for solving complex problems in various fields of science and engineering.
In this blog post, we will explore a range of JEE Mains questions related to work, power, and energy. These questions will help you enhance your understanding of these topics and prepare you for the challenging exam ahead. So, let’s dive in!
Overview: Work, Power, and Energy
Before we delve into the questions, let’s have a brief overview of the core concepts of work, power, and energy.
Work is the transfer of energy that occurs when a force acts upon an object and displaces it along the direction of the force. Mathematically, work is given by the formula:
Work = Force × Displacement × cos(θ)
where θ represents the angle between the applied force and the direction of displacement.
Power, on the other hand, measures how quickly work is done. It is defined as the rate at which work is done or energy is transferred. Mathematically, power can be calculated using the formula:
Power = Work/Time
Energy, in physics, represents the ability to do work, and it can exist in different forms such as kinetic energy, potential energy, and thermal energy. The law of conservation of energy states that energy can neither be created nor destroyed; it can only be transferred or transformed from one form to another.
JEE Mains Questions on Work, Power, and Energy
Now that we have refreshed our knowledge of work, power, and energy, let’s move on to the JEE Mains questions:
- A block of mass 2 kg is lifted vertically upwards by a force, such that its velocity remains constant. If the block is lifted by 2 meters in 5 seconds, calculate the work done by the force and the power exerted. (JEE Mains 2019)
- A car of mass 1000 kg is moving with a velocity of 20 m/s. If the car takes 10 seconds to stop after the application of the brakes, calculate its retardation and the distance covered during the stopping. (JEE Mains 2018)
- Two bodies of masses m1 and m2 (m2 > m1) are dropped from the same height. Calculate the ratio of their kinetic energies just before hitting the ground. (JEE Mains 2017)
- A machine does 400 J of work in 4 seconds. Calculate its power. (JEE Mains 2016)
- A body of mass 10 kg is thrown vertically upwards with an initial velocity of 20 m/s. Calculate its maximum height and the work done by gravity during its upward motion. (JEE Mains 2015)
Solutions and Detailed Answers
Now, let’s solve the questions one by one:
Question 1:
In this question, we are given:
| Mass of the block (m) | 2 kg |
|---|---|
| Vertical displacement (s) | 2 meters |
| Time taken (t) | 5 seconds |
To calculate the work done by the force, we can use the formula:
Work = Force × Displacement × cos(θ)
Since the velocity remains constant, the net force acting on the block is zero (as the force applied balances the force of gravity). Hence, the work done by the force is also zero.
To calculate the power exerted, we can use the formula:
Power = Work/Time
As the work done is zero, the power exerted is also zero.
Thus, the work done by the force is 0 J, and the power exerted is 0 W.
Question 2:
In this question, we are given:
| Mass of the car (m) | 1000 kg |
|---|---|
| Initial velocity (u) | 20 m/s |
| Time taken to stop (t) | 10 seconds |
To calculate the retardation, we can use the formula:
Retardation (a) = Change in velocity/Time taken
The change in velocity is given by:
Change in velocity = Final velocity – Initial velocity
As the car comes to a stop, the final velocity is zero. Substituting the given values, we get:
Retardation (a) = (0 – 20 m/s) / 10 s = -2 m/s2
To calculate the distance covered during the stopping, we can use the formula:
Distance (s) = (Initial velocity2 – Final velocity2) / (2 × Retardation)
Substituting the given values, we get:
Distance (s) = (202 – 02) / (2 × -2) = 200 meters
Thus, the retardation of the car is -2 m/s2, and the distance covered during the stopping is 200 meters.
Question 3:
In this question, we are given:
| Mass of body 1 (m1) | 10 kg |
|---|---|
| Mass of body 2 (m2) | 20 kg |
We are asked to calculate the ratio of the kinetic energies just before hitting the ground. Kinetic energy is given by the formula:
Kinetic energy = (1/2) × Mass × Velocity2
The velocity just before hitting the ground is the same for both bodies, as they are dropped from the same height and experience the same acceleration due to gravity. Therefore, the ratio of their kinetic energies is given by:
Ratio = (1/2) × m1 × (velocity)2 / (1/2) × m2 × (velocity)2
Simplifying the equation, we get:
Ratio = m1 / m2 = 10 kg / 20 kg = 1/2
Thus, the ratio of their kinetic energies just before hitting the ground is 1:2.
Question 4:
In this question, we are given:
| Work done (W) | 400 J |
|---|---|
| Time taken (t) | 4 seconds |
To calculate power, we can use the formula:
Power = Work/Time
Substituting the given values, we get:
Power = 400 J / 4 s = 100 W
Thus, the power exerted by the machine is 100 W.
Question 5:
In this question, we are given:
| Mass of the body (m) | 10 kg |
|---|---|
| Initial velocity (u) | 20 m/s |
Since the body is thrown upwards, we need to consider its motion in two parts: upward motion and downward motion.
To calculate the maximum height, we can use the formula:
Maximum height (h) = (Initial velocity2) / (2 × Acceleration due to gravity)
Substituting the given values, we get:
Maximum height (h) = (202) / (2 × 9.8) = 20.41 meters
To calculate the work done by gravity during the upward motion, we can use the formula:
Work = Force × Displacement × cos(θ)
Here, the force and displacement are in opposite directions, so the angle θ is 180 degrees. Thus, the work done by gravity is given by:
Work = (Force of gravity) × (Displacement) × cos(180 degrees)
Simplifying the equation, we get:
Work = -mgd
where d is the displacement and m is the mass.
Therefore, the work done by gravity during the upward motion is equal to the negative of the work done by the applied force, which is zero since the velocity remains constant.
Thus, the maximum height reached by the body is 20.41 meters, and the work done by gravity during the upward motion is 0 J.
Conclusion
In this blog post, we have explored a range of JEE Mains questions related to work, power, and energy. By solving these questions, we have enhanced our understanding of these fundamental topics. Remember to practice similar questions regularly to strengthen your problem-solving skills and increase your chances of success in the JEE Mains exam.
Resources:
- https://en.wikipedia.org/wiki/Work_(physics)
- https://en.wikipedia.org/wiki/Power_(physics)
- https://en.wikipedia.org/wiki/Energy
Happy learning and best of luck with your JEE Mains preparation!
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