Question
4. A 16.2 kg box is dropped straight down from a height of 4.0 m. Using the Law of conservation of energy determine the speed of the box the moment before it hits the ground (ignore any air resistance or friction). (4)
Solution
4.2
(329 Votes)
Maya
Elite · Tutor for 8 years
Answer
Here's how to solve this problem using the Law of Conservation of Energy:1. **Understand the principle:** The Law of Conservation of Energy states that energy cannot be created or destroyed, only transformed from one form to another. In this case, the box initially has potential energy due to its height, and this energy is converted into kinetic energy as it falls.2. **Set up the equation:** We can express the conservation of energy as: Potential Energy (initial) = Kinetic Energy (final) PE = KE3. **Define the formulas:** * Potential Energy (PE) = mgh, where m is the mass (16.2 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height (4.0 m). * Kinetic Energy (KE) = (1/2)mv², where m is the mass (16.2 kg) and v is the velocity (what we want to find).4. **Substitute and solve:** mgh = (1/2)mv² Notice that the mass (m) appears on both sides of the equation, so we can cancel it out: gh = (1/2)v² Now, plug in the values for g and h: (9.8 m/s²)(4.0 m) = (1/2)v² 39.2 m²/s² = (1/2)v² Multiply both sides by 2: 78.4 m²/s² = v² Take the square root of both sides to find v: v = √(78.4 m²/s²) v ≈ 8.85 m/s**Answer:** The speed of the box the moment before it hits the ground is approximately 8.85 m/s.