{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW16 - Pickering Tracey Homework 16 Due Mar 1 2006 noon...

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

Pickering, Tracey – Homework 16 – Due: Mar 1 2006, noon – Inst: Drummond 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A 1 kg mass slides to the right on a surface having a coefficient of friction 0 . 1 as shown in the figure. The mass has a speed of 7 m / s when contact is made with a spring that has a spring constant 55 N 2 . The mass comes to rest after the spring has been compressed a distance d . The mass is then forced toward the left by the spring and continues to move in that direction beyond the unstretched posi- tion. Finally the mass comes to rest a distance D to the left of the unstretched spring. The acceleration of gravity is 9 . 8 m / s 2 . m k v v v = 0 v = 0 D d i f Find the compressed distance d . Correct answer: 0 . 92623 m. Explanation: The principle we are going to use to solve this problem is that the change in the total energy of the system is equal to the work done by the nonconservative forces W nc = Δ K + Δ U . In our case, the nonconservative force is the frictional force. Therefore W nc = - f d = - (0 . 98 N) d , where the frictional force is f = μ m g = 0 . 1 (1 kg) (9 . 8 m / s 2 ) = 0 . 98 N . The change in kinetic energy is Δ K = 0 - 1 2 m v 2 i = - 1 2 (1 kg) (7 m / s) 2 = - 24 . 5 J , and the change in potential energy is Δ U = 1 2 k d 2 = 1 2 (55 N 2 ) d 2 = (27 . 5 N / m) d 2 , so W nc = Δ K + Δ U - (0 . 98 N) d = - 24 . 5 J + (27 . 5 N / m) d 2 . (27 . 5 N / m) d 2 + (0 . 98 N) d + ( - 24 . 5 J) = 0 . This is a quadratic equation, and since b 2 - 4 ac = (0 . 98 N) 2 - 4(27 . 5 N / m)( - 24 . 5 J) = 2695 . 96 N 2 , the positive solution is d = - (0 . 98 N) + 2695 . 96 N 2 2 (27 . 5 N / m) = 0 . 92623 m . 002 (part 1 of 3) 10 points A single conservative force F ( x ) = b x + a

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

View Full Document
Pickering, Tracey – Homework 16 – Due: Mar 1 2006, noon – Inst: Drummond 2 acts on a 2 . 35 kg particle, where x is in meters, b = 4 . 38 N / m and a = 1 . 87 N. As the particle moves along the x axis from x 1 = 0 . 601 m to x 2 = 4 . 29 m, calculate the work done by this force.
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