Lecture Notes: Work Done by a Varying Force

Introduction

• Topic: Work done by a varying force.
• Unlike constant force, varying force changes with displacement.
• Goal: Calculate work done and find the velocity at a particular point.

Basics of Work Done

• Standard Formula: Work done = Force x Displacement x cos(θ).
  • θ is the angle between force and displacement.
  • Valid only for constant forces.
• Problem: Force varies as displacement changes.

Problem Setup

• Given:
  • Mass of the object = 2kg.
  • Force applied from x = 0m to x = 5m.
  • Initial condition: Object at rest.
• Objective:
  • Find total work done.
  • Calculate velocity at x = 5m.

Calculating Work Done

• Method: Integration of force over displacement.
  • Formula: ( int F , dx ) where F is the instantaneous force.
  • Force expression: ( 5x^2 + 9x - 5 ).
• Integration Steps:
  • Use power rule for integration: ( int x^n , dx = frac{x^{n+1}}{n+1} ).
  • Compute:
    - ( 5x^3/3 + 9x^2/2 - 5x ).
  • Evaluate from 0 to 5.
• Result:
  • Total work done = 296 Joules.

Calculating Final Velocity

• Concept: Work-Energy Theorem.
  • Change in kinetic energy = Work done.
• Kinetic Energy:
  • ( KE_f - KE_i = ext{Work} ).
  • Initial KE = 0 (object starts from rest).
  • ( KE_f = rac{1}{2}mv^2 ).
• Calculation:
  • Mass = 2kg.
  • Solve for final velocity ( v_f ).
  • Final velocity = 17.2 m/s.

Key Concepts

• Integration for varying force.
• Work-Energy Theorem for calculating velocity.

Conclusion

• Two key concepts used:
  1. Integration for work done by varying force.
  2. Work-Energy Theorem for velocity.
• Reminder: For queries or feedback, comment below.

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