{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW 5 - johnson(mjj622 HW 5 Coker(56625 This print-out...

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

johnson (mjj622) – HW 5 – Coker – (56625) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This fairly difficult chapter (6) applies New- ton’s 2nd and 3rd Laws to examples with friction, with spring forces, and with circular motion at constant or variable speed. 001 10.0 points Two blocks are arranged at the ends of a mass- less string as shown in the figure. The system starts from rest. When the 2 . 92 kg mass has fallen through 0 . 377 m, its downward speed is 1 . 32 m / s. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 92 kg 4 . 59 kg μ a What is the frictional force between the 4 . 59 kg mass and the table? Correct answer: 11 . 2613 N. Explanation: Given : m 1 = 2 . 92 kg , m 2 = 4 . 59 kg , v 0 = 0 m / s , and v = 1 . 32 m / s . Basic Concept: Newton’s Second Law F = M a Solution: The acceleration of m 1 is obtained from the equation v 2 v 2 0 = 2 a ( s s 0 ) a = v 2 v 2 0 2 h = (1 . 32 m / s) 2 (0 m / s) 2 2 (0 . 377 m) = 2 . 31088 m / s 2 . Consider free body diagrams for the two masses T m 1 g a T N μ N a m 2 g Because m 1 and m 2 are tied together with string, they have same the speed and the same acceleration, so the net force exerted on m 2 is F 2 = m 2 a The net force on m 1 is m 1 a = m 1 g T , so that T = m 1 g m 1 a . Thus F 2 = T f k , f k = T F 2 = m 1 g ( m 1 + m 2 ) a = (2 . 92 kg) (9 . 8 m / s 2 ) (2 . 92 kg + 4 . 59 kg) × (2 . 31088 m / s 2 ) = 11 . 2613 N . 002 10.0 points A block of mass 7 . 8 kg rests on a plane inclined at an angle of 27 . The static coefficient of friction between the block and the plane is 0 . 63. What is the frictional force on the block? 1. 42 . 9084 N 2. 68 . 1085 N 3. 8 . 20535 N 4. 34 . 703 N correct 5. 21 . 8629 N

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

View Full Document
johnson (mjj622) – HW 5 – Coker – (56625) 2 Explanation: The maximum static friction is f max = mgμ cos θ = (7 . 8 kg)(9 . 8 m / s 2 )(0 . 63) cos 27 = 42 . 9084 N Since mg sin θ = (7 . 8 kg)(9 . 8 m / s 2 ) sin 27 = 34 . 703 N the friction is large enough to keep the block from moving, so the static friction is equal to mg sin θ . 003 10.0 points An object is held in place by friction on an inclined surface. The angle of inclination is increased until the object starts moving. If the surface is kept at this angle, the object 1. slows down. 2. speeds up. correct 3. moves at uniform speed. 4. none of the other choices Explanation: As the tilt of the surface is increased at a certain angle the object starts sliding. Until that angle is reached, the object is at rest and the net force on it is zero. For the object to start sliding from rest, there must be a net force on it; if the net force on it is no longer zero, the object will accelerate. 004 (part 1 of 2) 10.0 points A block of mass m is accelerated across a rough surface by a force of magnitude F that is exerted at an angle φ with the horizontal, as shown above. The frictional force on the block exerted by the surface has magnitude f .
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