A block of mass 2 kg is sliding down a frictionless incline of height 5 m. What is its speed at the bottom of the incline?
Practice Questions
1 question
Q1
A block of mass 2 kg is sliding down a frictionless incline of height 5 m. What is its speed at the bottom of the incline?
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. Thus, v = sqrt(2gh) = sqrt(2 * 9.81 * 5) = 10 m/s.
Questions & Step-by-step Solutions
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Q
Q: A block of mass 2 kg is sliding down a frictionless incline of height 5 m. What is its speed at the bottom of the incline?
Solution: Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Thus, v = sqrt(2gh) = sqrt(2 * 9.81 * 5) = 10 m/s.
