Work Done by Constant and Variable Forces: Practice Problem & Solution
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)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Work Done by Constant and Variable Forces. Understanding the underlying formula is key to arriving at the correct answer below:
\(W = int F dx\). Given \(F = 7 - 2x + 3x^2\). Limits \(x = 0\) to \(x = 5\text{ m})\). \(W = [7x - x^2 + x^3]_0^5 = 7(5) - 5^2 + 5^3 = 35 - 25 + 125 = 135\text{ J}\).
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