A 10 kg block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (2020)
Practice Questions
1 question
Q1
A 10 kg block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (2020)
5 m/s
10 m/s
15 m/s
20 m/s
Using conservation of energy, PE at top = KE at bottom. mgh = 1/2 mv^2. 10 kg * 9.8 m/s^2 * 5 m = 1/2 * 10 kg * v^2. Solving gives v = 10 m/s.
Questions & Step-by-step Solutions
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Q
Q: A 10 kg block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (2020)
Solution: Using conservation of energy, PE at top = KE at bottom. mgh = 1/2 mv^2. 10 kg * 9.8 m/s^2 * 5 m = 1/2 * 10 kg * v^2. Solving gives v = 10 m/s.
Steps: 11
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: 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 = 10 kg * 9.8 m/s² * 5 m.
Step 5: Calculate the potential energy: PE = 490 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: 490 Joules = 1/2 * 10 kg * v².
Step 8: Rearrange the equation to solve for v²: v² = (2 * 490 Joules) / 10 kg.
Step 9: Calculate v²: v² = 98 m²/s².
Step 10: Take the square root of v² to find v: v = √98 m²/s².
Step 11: Calculate the final speed: v ≈ 9.9 m/s, which can be rounded to 10 m/s.
