A block of mass 2 kg is released from a height of 10 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 10 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
14 m/s
20 m/s
10 m/s
5 m/s
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 10) = 14 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 10 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 = sqrt(2gh) = sqrt(2 * 9.8 * 10) = 14 m/s.
Steps: 9
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 10 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 * 10 m = 196 Joules.
Step 6: Use the formula for kinetic energy (KE) just before hitting the ground: KE = 0.5 * m * v², where v is the speed.
Step 7: Set the potential energy equal to the kinetic energy: 196 Joules = 0.5 * 2 kg * v².
Step 8: Simplify the equation: 196 = 1 * v², so v² = 196.
Step 9: Take the square root of both sides to find v: v = sqrt(196) = 14 m/s.
