Work, Energy & Power L2 | Work Done by a Constant & Variable Force | Physics Class 11, JEE, NEET

Introduction

This tutorial covers the concepts of work done by constant and variable forces, as discussed in the physics class video. Understanding these concepts is essential for students preparing for Class 11 physics, JEE, and NEET exams. This guide will break down the key principles and provide practice questions to reinforce learning.

Step 1: Understanding Work Done by Constant Force

• Definition of Work: Work is defined as the product of force and the distance moved in the direction of the force.

• Formula:

  • For a constant force: [ W = F \cdot d \cdot \cos(\theta) ] Where:
    • (W) is work done,
    • (F) is the force applied,
    • (d) is the distance moved,
    • (\theta) is the angle between the force and the direction of motion.

• Practical Advice:

  • When calculating work done, ensure that the angle (\theta) is correctly identified. For instance, if the force is in the same direction as the motion, (\theta = 0) degrees, making (\cos(0) = 1).

Step 2: Calculating Work Done Against Gravity

• Work Done by Gravity:

  • When an object moves vertically, the work done by gravity can be calculated using: [ W = m \cdot g \cdot h ] Where:
    • (m) is the mass of the object,
    • (g) is the acceleration due to gravity (approximately (9.81 , \text{m/s}^2)),
    • (h) is the height moved.

• Common Pitfalls:

  • Be cautious with signs. Work done against gravity is considered positive when lifting an object.

Step 3: Understanding Work Done by Variable Force

• Definition: A variable force is one that changes in magnitude or direction.

• Work Done by Variable Force:

  • The work done can be calculated using calculus: [ W = \int F(x) , dx ] Where (F(x)) is the force as a function of position.

• Practical Advice:

  • To find the total work done over a distance, determine the function (F(x)) and integrate over the specified limits.

Step 4: Practice Questions to Reinforce Learning

• Question 1: A force of 10 N is applied to move an object 5 m in the direction of the force. Calculate the work done.

  • Solution: [ W = 10 , \text{N} \cdot 5 , \text{m} = 50 , \text{J} ]

• Question 2: Calculate the work done by a variable force described by the equation (F(x) = 3x^2) from (x = 0) to (x = 2).

  • Solution: [ W = \int_{0}^{2} 3x^2 , dx = [x^3]_{0}^{2} = 2^3 - 0^3 = 8 , \text{J} ]

Conclusion

Understanding work done by constant and variable forces is crucial in physics. This guide provided you with definitions, formulas, and practice questions to enhance your grasp of these concepts. For further study, consider exploring related topics such as energy conservation and power calculations.