# OLDHW16 - oldhomewk 16 PAPAGEORGE MATT Due 4:00 am Question...

• Notes
• 5

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

oldhomewk 16 – PAPAGEORGE, MATT – Due: Feb 24 2008, 4:00 am 1 Question 1, chap 7, sect 2. part 1 of 2 10 points A single conservative force acting on a par- ticle varies as vector F = ( - A x + B x 2 ) ˆ ı , where A = 40 N / m and B = 84 N / m 2 and x is in meters. Find the change in potential energy as the particle moves from x 0 = 3 . 2 m to x 1 = 2 . 8 m . Correct answer: 254 . 848 J (tolerance ± 1 %). Explanation: The potential energy is U ( x ) = - integraldisplay x 0 ( - A x + B x 2 ) d x = A x 2 2 - B x 3 3 . If we take U (0) = 0, then the change in the potential energy is U = U ( x 1 ) - U ( x 0 ) = parenleftbigg Ax 2 1 2 - Bx 3 1 3 parenrightbigg - parenleftbigg Ax 2 0 2 - Bx 3 0 3 parenrightbigg = bracketleftbigg (40 N / m) (2 . 8 m) 2 2 - (84 N / m 2 ) (2 . 8 m) 3 3 bracketrightbigg - bracketleftbigg (40 N / m) (3 . 2 m) 2 2 - (84 N / m 2 ) (3 . 2 m) 3 3 bracketrightbigg = 254 . 848 J . Question 2, chap 7, sect 2. part 2 of 2 10 points Find the change in kinetic energy of the particle between the same two points. Correct answer: - 254 . 848 J (tolerance ± 1 %). Explanation: From conservation of energy (conservative force), the change of the kinetic energy is K = - U = - (254 . 848 J) = - 254 . 848 J . Question 3, chap 7, sect 1. part 1 of 2 10 points Starting from rest at a height equal to the radius of the circular track, a block of mass 26 kg slides down a quarter circular track under the influence of gravity with friction present (of coefficient μ ). The radius of the track is 30 m. The acceleration of gravity is 9 . 8 m / s 2 . 30 m 26 kg θ Determine the work done by the conserva- tive forces. Correct answer: 7644 J (tolerance ± 1 %). Explanation: The work done by the conservative force (gravity) is W grav = m g R = (26 kg) (9 . 8 m / s 2 ) (30 m) = 7644 J . Question 4, chap 7, sect 1. part 2 of 2 10 points If the kinetic energy of the block at the bottom of the track is 5400 J, what is the work done against friction? Correct answer: 2244 J (tolerance ± 1 %). Explanation: W = W f + K W f = m g R - K = (26 kg) (9 . 8 m / s 2 ) (30 m) - (5400 J) = 2244 J .

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

oldhomewk 16 – PAPAGEORGE, MATT – Due: Feb 24 2008, 4:00 am 2 Question 5, chap 8, sect 1.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern