{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Practice_Test_5_Answers - 1 Solutions to Practice Test 5 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Solutions to Practice Test 5 1. A mass hung from the end of a spring is executing simple harmonic motion. Where is the potential energy of the oscillating mass a maximum? Potential energy is a maximum at positions where the spring is compressed or stretched to the maximum. This happens at the two extreme positions. b. At the two extreme positions 2. Where is the kinetic energy of the oscillating mass in question 1 a maximum? a. At the equilibrium position 3. The total energy of the oscillating object in question 1 is Although energy changes from potential to kinetic and back again, the total energy remains a constant at all positions. d. the same at all points 4. The acceleration of the oscillating mass in question 1 is Acceleration is a maximum at positions where the restoring force is a maximum. This happens at the extreme positions. c. a maximum at the extreme positions 5. The restoring force on the oscillating mass is a. always in a direction opposite to the displacement 6. The restoring force on the oscillating mass is c. directly proportional to the displacement 7. A 50 g mass is suspended from the free end of a spring of spring constant 400 N/m produces a displacement of 8.2 cm. What is the work done on the spring? 2 2 1 1 40 0.082 0.13 J 2 2 W kx = = × × = a. 0.13 J 8. What is the potential energy of the spring in question 7? Potential energy is a measure of the work done to stretch it. a. 0.13 J 9. In question 7, if the mass is pulled down by a distance of 5 cm and released, what will be its kinetic energy when it returns to the equilibrium position? The mass will oscillate with an amplitude of 0.05 m. The potential energy at the lower extreme = ½ kA 2 = ½ × 40 × 0.05 2 = 0.05 J a. 0.05 J 10. What is the maximum velocity attained by the mass in question 9? All the potential energy at the lower extreme becomes potential energy at the equilibrium position. The maximum velocity occurs here.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 2 max 2 max -1 max 1 0.05 2 2 0.05 0.1 2 0.05 1.41 ms mv v m v = × = = = = a. 1.41 m/s 11. What will be the kinetic energy of the mass in question 9 when it is at the upper extreme? At the upper extreme, all the energy is potential, kinetic energy = 0 c. 0 J 12. What is the period of a 1.0-m long pendulum where g = 9.8 m/s 2 ? Period of a simple pendulum is: 1 2 2 2.0 s 9.8 L T g π π = = = b. 2.0 s 13. What is the length of a pendulum on the surface of the moon if its period on the moon is 4.8 s? (g on the moon is 1.63 m/s 2 ) 2 2 2 2 4.8 2 1.63 4.8 4 1.63 4.8 1.63 0.95 m 4 L L L π π π = = × = = b. 0.95 m 14.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}