A 2 kg block is sliding down a frictionless incline of height 5 m. What is its speed at the bottom?
Practice Questions
1 question
Q1
A 2 kg block is sliding down a frictionless incline of height 5 m. What is its speed at the bottom?
10 m/s
5 m/s
20 m/s
15 m/s
Using conservation of energy, potential energy at the top = kinetic energy at the bottom: 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 2 kg block is sliding down a frictionless incline of height 5 m. What is its speed at the bottom?
Solution: Using conservation of energy, potential energy at the top = kinetic energy at the bottom: mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 5) = 10 m/s.
Steps: 10
Step 1: Identify the mass of the block, which is 2 kg.
Step 2: Identify the height of the incline, which is 5 m.
Step 3: Use the formula for gravitational potential energy (PE) at the top: PE = mgh, where g is the acceleration due to gravity (approximately 9.8 m/s²).
Step 4: Calculate the potential energy at the top: PE = 2 kg * 9.8 m/s² * 5 m.
Step 5: Calculate the potential energy: PE = 2 * 9.8 * 5 = 98 Joules.
Step 6: At the bottom of the incline, all potential energy converts to kinetic energy (KE). The formula for kinetic energy is KE = 0.5 * mv².
Step 7: Set the potential energy equal to the kinetic energy: 98 Joules = 0.5 * 2 kg * v².
Step 8: Simplify the equation: 98 = 1 * v², so v² = 98.
Step 9: Take the square root of both sides to find v: v = √98.
Step 10: Calculate the speed: v ≈ 9.9 m/s, which can be rounded to 10 m/s.
