Work Done by Constant and Variable Forces: Practice Problem & Solution
A force \(F = 20 + 10y\) acts on a particle in \(y\) direction where \(F\) is in newton and \(y\) in meter. Work done by this force to move the particle from \(y = 0\) to \(y = 1\text{ m})\) is: (2019)
Solution Explained:
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:
Work \(W = \int F dy\). Given \(F = 20 + 10y\). Limits \(y = 0\) to \(y = 1\text{ m})\). \(W = \int_0^1 (20 + 10y) dy = [20y + 5y^2]_0^1 = 20(1) + 5(1)^2 = 25\text{ J}\).
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