OMPhQuiz_07_p58-66 - chapter 7 Work and Energy Work Work in...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
58 PhysiQuiz Contents Work Work in one dimension: (Section 7.1, part 1) The following two problems involve pulleys and may be used after Example 2 in Section 7.1: 1. Atwood machine 2. Raising m by a distance D x Work in two dimensions: (Section 7.1, part 2) 3. Work against gravity • Context in the textbook: After Example 3 in Section 7.1. Work–energy and gravitational potential energy (Section 7.4) Exercises to be used in conjunction with Section 7.4: 4. Sliding down an incline 5. Comparing final kinetic energies 6. Stopped pendulum 7. Sliding down a dome: The “stay-on” condition 8. Sliding down a dome: Equations of motion chapter 7 Work and Energy
Background image of page 1

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

View Full DocumentRight Arrow Icon
Two blocks with masses m 1 and m 2 are connected by a light string passing over a light frictionless pulley. Assume m 2 . m 1 . Determine the potential energy released by the system as block 2 starts from rest and falls by h /2: AB C Potential energy released m 2 gh /2 ( m 2 2 m 1 ) gh /2 m 1 gh /2 Hint: Remember that as block 2 descends by h /2, block 1 rises correspondingly by a height h /2. Explanation: When block 2 falls by a distance h /2, block 1 goes up by h /2. The potential energy lost by block 2 is m 2 gh /2. The potential energy gained by block 1 is m 1 gh /2. So the net potential energy lost is ( m 2 2 m 1 ) gh /2. Answer 5 B. Block 1 Block 2 m 1 m 2 h 1. Atwood Machine PhysiQuiz 59
Background image of page 2
Consider the mass–pulley system shown. Determine the distance d covered by the force F as it lifts the block by a height D x : AB C Distance of F d 5D x /2 d xd 5 2 D x Explanation: As the block is lifted by height D x , the length of each of the two strings supporting the moving pulley will be reduced by D x , so d 5 2 D x . Answer 5 C. Comment: Notice that conservation of energy implies that the increase of the potential energy equals the input mechanical energy: mg D x 5 F d 5 2 D x F. Solving for F gives F 5 mg /2. In other words, the applied force required is half of the weight.
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 9

OMPhQuiz_07_p58-66 - chapter 7 Work and Energy Work Work in...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online