Work Done by Position Dependent Force – Rankers Physics
Topic: Work Energy and Power
Subtopic: Work Done by Constant and Variable Forces

Work Done by Position Dependent Force

A position dependent force, \(F = (7 - 2x + 3x^2)\text{ N})\) acts on a small body of mass \(2\text{ kg})\) and displaces it from \(x = 0\) to \(x = 5\text{ m})\). The work done in joule is:

(1994, 92)

135
270
35
70

Solution:

\(W = int F dx\). Given \(F = 7 - 2x + 3x^2\). Limits \(x = 0\) to \(x = 5\text{ m})\). \(W = [7x - x^2 + x^3]_0^5 = 7(5) - 5^2 + 5^3 = 35 - 25 + 125 = 135\text{ J}\).

Leave a Reply

Your email address will not be published. Required fields are marked *