Solution:
The increase in potential energy is \(\Delta U = \frac{mgh}{1 + h/R}\). Since \(h = nR\), we get \(\Delta U = \frac{mg(nR)}{1 + n} = \left(\frac{n}{n+1}\right) mgR\).
The increase in potential energy is \(\Delta U = \frac{mgh}{1 + h/R}\). Since \(h = nR\), we get \(\Delta U = \frac{mg(nR)}{1 + n} = \left(\frac{n}{n+1}\right) mgR\).
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