Principal of Conservation of Energy: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Principal of Conservation of Energy. Understanding the underlying formula is key to arriving at the correct answer below:

Maximum speed occurs at minimum height, where potential energy is lowest and kinetic energy is highest. Minimum speed (zero) occurs at maximum height. By conservation of mechanical energy: \(mgh_{max} + \frac{1}{2}mv_{min}^2 = mgh_{min} + \frac{1}{2}mv_{max}^2\). With \(v_{min}=0\), \(mg(2) = mg(0.75) + \frac{1}{2}mv_{max}^2\). \(2g - 0.75g = \frac{1}{2}v_{max}^2 \Rightarrow 1.25g = \frac{1}{2}v_{max}^2\). Using \(g = 10\text{ m/s}^2\), \(v_{max}^2 = 2.5 \times 10 = 25\), so \(v_{max} = 5\text{ m/s}\).