{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solution22_pdf - arellano(aa39398 Practice Homework Chapter...

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

View Full Document Right Arrow Icon
arellano (aa39398) – Practice Homework Chapter 8 Solutions – weathers – (17101) 1 This print-out should have 58 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A 42.0 kg child is in a swing that is attached to ropes 2.30 m long. The acceleration of gravity is 9 . 81 m / s 2 . Find the gravitational potential energy as- sociated with the child relative to the child’s lowest position under the following condi- tions: a) when the ropes are horizontal. Correct answer: 947 . 646 J. Explanation: Basic Concept: U g = mgh Given: = 2 . 30 m m = 42 . 0 kg g = 9 . 81 m / s 2 Solution: The child is at a height h 1 = from the lowest point, so U g = mgh 1 = (42 kg)(9 . 81 m / s 2 )(2 . 3 m) = 947 . 646 J 002 (part 2 of 3) 10.0 points b) when the ropes make a 33 . 0 angle with the vertical. Correct answer: 152 . 883 J. Explanation: Given: θ = 33 . 0 Solution: The child is a distance cos θ from the at- tachment point, so h 2 = cos θ and U g = mgh 2 = mgℓ (1 cos θ ) = (42 kg)(9 . 81 m / s 2 )(2 . 3 m) · (1 cos 33 ) = 152 . 883 J 003 (part 3 of 3) 10.0 points c) at the bottom of the circular arc. Correct answer: 0 J. Explanation: Solution: The child is at the lowest point, so h 3 = 0, and U g = mgh 3 = (42 kg)(9 . 81 m / s 2 )(0 m) = 0 J 004 (part 1 of 3) 10.0 points A 2 . 94 kg ball is attached to a ceiling by a 1 . 88 m long string. The height of the room is 5 . 05 m . The acceleration of gravity is 9 . 8 m / s 2 . What is the gravitational potential energy associated with the ball relative to the ceiling? Correct answer: 54 . 1666 J. Explanation: Given : m = 2 . 94 kg and Δ h 1 = 1 . 88 m . U g = m g Δ h 1 = (2 . 94 kg) ( 9 . 8 m / s 2 ) ( 1 . 88 m) = 54 . 1666 J . 005 (part 2 of 3) 10.0 points What is its gravitational potential energy rel- ative to the floor? Correct answer: 91 . 334 J. Explanation: Let : h 2 = 5 . 05 m 1 . 88 m = 3 . 17 m U g = mgh 2 = (2 . 94 kg) ( 9 . 8 m / s 2 ) (3 . 17 m) = 91 . 334 J .
Image of page 1

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

View Full Document Right Arrow Icon
arellano (aa39398) – Practice Homework Chapter 8 Solutions – weathers – (17101) 2 006 (part 3 of 3) 10.0 points What is its gravitational potential energy rel- ative to a point at the same elevation as the ball? Correct answer: 0 J. Explanation: Given : h 3 = 0 m U g = mgh 3 = (2 . 94 kg) ( 9 . 8 m / s 2 ) (0 m) = 0 J . 007 10.0 points A synthetic rubber band resists being stretched a distance x from equilibrium with a force vector F b ( x ) = ˆ ı b x 2 , where b is a constant. What is the potential energy U b ( x ) associ- ated with this elastic band? 1. b x 3 3 2. zero 3. 2 b x 4. + b x 2 2 5. + b x 3 3 correct 6. +2 b x 7. b x 2 2 Explanation: By definition Δ U = W . The work we would have to do against the force vector F b to stretch the band is integraldisplay b x 2 dx = b x 3 3 . 008 10.0 points A ball of mass m is hung from the end of a vertical, unstretched spring of negligible mass and force constant k , then released from rest. After it has fallen a vertical distance it is momentarily at rest again.
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 ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern