A 2 kg ball is dropped from 5 m. Using energy conservation, what is its speed just before hitting the ground? (Use g = 10 m/s²)
A5 m/s
B10 m/s
C20 m/s
D50 m/s
PE at top = KE at bottom: mgh = ½mv². The mass cancels: gh = ½v². So v² = 2gh = 2 × 10 × 5 = 100. v = 10 m/s.
Question 2 True / False
If a roller coaster has 50,000 J of potential energy at the top of a hill and no kinetic energy, it will have 50,000 J of kinetic energy at the bottom (ignoring friction).
TTrue
FFalse
Answer: True
By conservation of energy, all the PE at the top converts to KE at the bottom when friction is ignored. Total energy stays at 50,000 J.
Question 3 Short Answer
A skateboarder rolls down a 3 m ramp starting from rest. What is their speed at the bottom? (Use g = 10 m/s²)
Think about your answer, then reveal below.
Model answer: About 7.7 m/s. Using mgh = ½mv², the mass cancels, giving v = √(2 × 10 × 3) = √60 ≈ 7.7 m/s.
Setting PE at top equal to KE at bottom: mgh = ½mv². Mass cancels. v = √(2gh) = √(2 × 10 × 3) = √60 ≈ 7.7 m/s.