AP Physics 1hardmcq1 pt
A pendulum bob passes through equilibrium with speed 2.0 m/s and swings 0.20 m high (g = 10 m/s²). Its speed at 0.10 m height is about
A.1.4 m/s
B.2.0 m/s
C.1.0 m/s
D.0 m/s
Explanation
Core Concept
½mv² = ½m(2)² − mgh: v = √(4 − 2·10·0.10) = √2 ≈ 1.4 m/s.
Correct Answer
A1.4 m/s
More Unit 6: Simple Harmonic Motion practice questions
- mediumA 0.40 kg mass on a spring oscillates with period 0.80 s. The spring constant is about (π² ≈ 10)
- easyA spring-block oscillator's restoring force at displacement +x is −kx. The acceleration is
- mediumA 2.0 kg mass on a spring of k = 50 N/m oscillates. Its frequency is about (π ≈ 3.14)
- easyA 1.0 kg mass on a spring of k = 100 N/m oscillates with amplitude 0.20 m. Its total energy is
- hardA 0.30 kg mass oscillates on a spring with amplitude 0.20 m. At displacement 0.10 m, its speed is 3.0 m/s. The spring constant is about
Practice more AP Physics 1 questions with full explanations
Practice Unit 6: Simple Harmonic Motion Questions →