Lecture-2_Work done by variable force and spring force by RKH SIR(480P)

Work Done by Variable Force

Introduction

  • Work done by a variable force involves calculating the work done along a displacement under a force that may change in magnitude or direction.

Basic Definition

  • Work done (W) can be defined as the dot product of the force (F) and a small displacement (ds):

    W = F · ds

  • This formula implies integrating the force over the displacement if the force varies.

Mathematical Representation

  • Work done by a force F over a displacement s is given by the integral:

    W = ∫ F · ds

  • The limits of integration correspond to the initial and final positions of the displacement.

Coordinate System

  • A Cartesian coordinate system can be used for representation:

    F = F_x î + F_y ĵ + F_z k̂

  • The displacement vector can be expressed as:

    ds = dx î + dy ĵ + dz k̂

Work Done in Component Form

  • Expressing work done in its component forms:

    W = ∫ (F_x dx + F_y dy + F_z dz)

  • Each component of the force is integrated separately over its respective displacement.

Important Considerations

  • Identify the limits for integration:

    • For x: from initial to final position (x1 to x2)

    • For y and z: similarly defined limits.

Example Calculations

  1. Example 1 (Particle moving under defined force):

    • Given F = x² î + y ĵ. If a particle moves from (1, 0, 1) to (2, 1, 3):

      W = ∫[1 to 2] (x² dx + y dy + 0 dz).

      • This simplifies to:W = (1/3) * 2³ - (1/3) * 1³ + (1/2) * (1 - 0).

      • Result: 15/6 Joules.

  2. Example 2 (Integrating using time as a parameter):

  • Given the force vector in terms of time: F = t² î + t ĵ over the interval (0 to 2 secs):

    • Calculate: W = ∫ (t² * 2 dt - 2t) from 0 to 2.

    • After integrating and applying limits:

    • Result: Zero (meaning no work done).

  1. Work Done by Spring Force:

  • The work done by a spring force when extending or compressing can be calculated as:

    W = -∫ kx dx from x1 to x2

  • Simplifying this yields:

    W = -1/2 k (x2² - x1²)

Summary

  • The key to solving problems related to work done by a variable force lies in accurately defining the force components, determining proper limits for integration, and applying calculus principles to evaluate the integral for work done, both in simple movements and complex scenarios like systems involving springs.