This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: morris (gmm643) – HW 5 – Coker – (56625) 1 This printout should have 19 questions. Multiplechoice 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 3 . 26 kg mass has fallen through 0 . 386 m, its downward speed is 1 . 28 m / s. The acceleration of gravity is 9 . 8 m / s 2 . 3 . 26 kg 4 . 52 kg μ a What is the frictional force between the 4 . 52 kg mass and the table? Correct answer: 15 . 4367 N. Explanation: Given : m 1 = 3 . 26 kg , m 2 = 4 . 52 kg , v = 0 m / s , and v = 1 . 28 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 = 2 a ( s − s ) a = v 2 − v 2 2 h = (1 . 28 m / s) 2 − (0 m / s) 2 2 (0 . 386 m) = 2 . 12228 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 = (3 . 26 kg) (9 . 8 m / s 2 ) − (3 . 26 kg + 4 . 52 kg) × (2 . 12228 m / s 2 ) = 15 . 4367 N . 002 10.0 points A block of mass 8 . 8 kg rests on a plane inclined at an angle of 26 ◦ . The static coefficient of friction between the block and the plane is . 64. What is the frictional force on the block? 1. 11 . 8026 N 2. 49 . 6077 N 3. 37 . 8051 N correct 4. 24 . 1953 N 5. 77 . 512 N morris (gmm643) – HW 5 – Coker – (56625) 2 Explanation: The maximum static friction is f max = mgμ cos θ = (8 . 8 kg)(9 . 8 m / s 2 )(0 . 64) cos26 ◦ = 49 . 6077 N Since mg sin θ = (8 . 8 kg)(9 . 8 m / s 2 ) sin 26 ◦ = 37 . 8051 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. speeds up. correct 2. slows down. 3. none of the other choices 4. moves at uniform speed. 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....
View
Full Document
 Spring '10
 COKER
 Force, Friction, Cos, Coker

Click to edit the document details