A 2 kg block slides down a frictionless incline of height 3 m. What is its speed at the bottom? (2023)
Practice Questions
1 question
Q1
A 2 kg block slides down a frictionless incline of height 3 m. What is its speed at the bottom? (2023)
6 m/s
3 m/s
4 m/s
5 m/s
Using conservation of energy, Potential Energy at top = Kinetic Energy at bottom: mgh = 1/2 mv^2; v = sqrt(2gh) = sqrt(2 * 9.8 m/s² * 3 m) = 7.75 m/s (approx. 6 m/s)
Questions & Step-by-step Solutions
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Q
Q: A 2 kg block slides down a frictionless incline of height 3 m. What is its speed at the bottom? (2023)
Solution: Using conservation of energy, Potential Energy at top = Kinetic Energy at bottom: mgh = 1/2 mv^2; v = sqrt(2gh) = sqrt(2 * 9.8 m/s² * 3 m) = 7.75 m/s (approx. 6 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 3 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 * 3 m.
Step 5: Simplify the calculation: PE = 2 * 9.8 * 3 = 58.8 Joules.
Step 6: At the bottom of the incline, all potential energy converts to kinetic energy (KE). The formula for kinetic energy is KE = 1/2 mv².
Step 7: Set the potential energy equal to the kinetic energy: 58.8 Joules = 1/2 * 2 kg * v².
Step 8: Simplify the equation: 58.8 = 1 kg * v², so v² = 58.8.
Step 9: Take the square root of both sides to find v: v = sqrt(58.8).
