A 10 kg block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (g = 10 m/s²) (2020)

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Q: A 10 kg block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (g = 10 m/s²) (2020)

Solution: Using conservation of energy, PE at top = KE at bottom: mgh = 1/2 mv²; v = sqrt(2gh) = sqrt(2 * 10 m/s² * 5 m) = 10 m/s

Steps: 10

Step 1: Identify the mass of the block, which is 10 kg.
Step 2: Identify the height of the incline, which is 5 m.
Step 3: Identify the acceleration due to gravity, which is 10 m/s².
Step 4: Use the formula for potential energy (PE) at the top of the incline: PE = mgh.
Step 5: Calculate the potential energy: PE = 10 kg * 10 m/s² * 5 m = 500 J.
Step 6: At the bottom of the incline, all potential energy converts to kinetic energy (KE).
Step 7: Use the formula for kinetic energy: KE = 1/2 mv².
Step 8: Set the potential energy equal to the kinetic energy: 500 J = 1/2 * 10 kg * v².
Step 9: Solve for v²: 500 J = 5 kg * v², so v² = 500 J / 5 kg = 100 m²/s².
Step 10: Take the square root to find v: v = sqrt(100 m²/s²) = 10 m/s.