In the context of a frictionless slide, how does increasing the starting height of an object affect its final speed at the bottom, assuming it starts from rest?

• A. The final speed decreases because the object takes longer to reach the bottom.
• B. The final speed increases because the object has more gravitational potential energy $E=mgh$ to convert into kinetic energy $E=\frac{1}{2}m{v}^2$.
• C. The final speed remains the same because mass cancels out in the energy equations.
• D. The final speed is independent of the starting height and depends only on the mass of the object.

Answer: (B)

Verified step by step guidance

1. Identify the forms of energy involved: at the top of the slide, the object has gravitational potential energy given by \(E = m g h\), where \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is the height.
2. Recognize that the slide is frictionless, so mechanical energy is conserved. This means the gravitational potential energy at the top converts entirely into kinetic energy at the bottom.
3. Write the kinetic energy at the bottom of the slide as \(E = \frac{1}{2} m v^2\), where \(v\) is the final speed of the object.
4. Set the initial potential energy equal to the final kinetic energy to relate height and speed: \(m g h = \frac{1}{2} m v^2\).
5. Solve this equation for the final speed \(v\) by canceling mass \(m\) and rearranging to get \(v = \sqrt{2 g h}\), showing that the final speed increases with the square root of the starting height.