A block of mass 2 kg is released from a height of 5 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
Practice Questions
1 question
Q1
A block of mass 2 kg is released from a height of 5 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
5 m/s
10 m/s
15 m/s
20 m/s
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 5) = 10 m/s.
Questions & Step-by-step Solutions
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Q
Q: A block of mass 2 kg is released from a height of 5 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
Solution: Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 5) = 10 m/s.
Steps: 14
Step 1: Identify the mass of the block, which is 2 kg.
Step 2: Identify the height from which the block is released, which is 5 m.
Step 3: Identify the acceleration due to gravity, which is 9.8 m/s².
Step 4: Use the formula for potential energy (PE) at height: PE = mgh, where m is mass, g is gravity, and h is height.
Step 5: Calculate the potential energy: PE = 2 kg * 9.8 m/s² * 5 m.
Step 6: Calculate the potential energy: PE = 98 Joules.
Step 7: Use the formula for kinetic energy (KE) just before hitting the ground: KE = 0.5mv².
Step 8: Set the potential energy equal to the kinetic energy: mgh = 0.5mv².
Step 9: Since mass (m) is on both sides, it can be canceled out: gh = 0.5v².
Step 10: Rearrange the equation to solve for v: v² = 2gh.
Step 11: Substitute the values of g and h into the equation: v² = 2 * 9.8 m/s² * 5 m.
Step 12: Calculate v²: v² = 98 m²/s².
Step 13: Take the square root to find v: v = √(98) m/s.
Step 14: Calculate the final speed: v ≈ 9.9 m/s, which can be rounded to 10 m/s.
