Uncategorized: Practice Problem & Solution
A man weighs \(80text{ kg}\) . He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of \(5text{ m/s}^2\). What would be the reading on the scale? \((g = 10text{ m/s}^2)\): (2003)
Solution Explained:
To solve this problem, we apply the core principles of Uncategorized. Understanding the underlying formula is key to arriving at the correct answer below:
When a lift accelerates upwards, the apparent weight \(R\) is given by \(R = m(g+a)\). Given \(m = 80text{ kg}\), \(g = 10text{ m/s}^2\), and \(a = 5text{ m/s}^2\). Therefore, \(R = 80text{ kg} times (10text{ m/s}^2 + 5text{ m/s}^2) = 80 times 15 = 1200text{ N}\).
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