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Unformatted text preview: reddy (ar38357) Friction clancy (SCI411-2) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (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 with the horizontal, as shown above. The frictional force on the block exerted by the surface has magnitude f . f F m What is the magnitude of the acceleration vectora of the block? 1. | vectora | = F m 2. | vectora | = F sin - m g m 3. | vectora | = F cos - f m correct 4. | vectora | = F cos m 5. | vectora | = F- f m Explanation: From Newtons second law of motion, the acceleration is the total force in the horizontal direction divided by the mass. There are two forces in the horizontal direction: one is the friction force; the other is the horizontal com- ponent of the dragging force F , but they are in the opposite directions, so the acceleration of the block is | vectora | = F cos - f m . 002 (part 2 of 2) 10.0 points Which of the following expressions for the coefficient of friction is correct? 1. = m g f 2. = f m g- F cos 3. = m g- F cos f 4. = f m g- F sin correct 5. = f m g Explanation: By definition, the coefficient of kinetic fric- tion is the ratio of the friction force and the normal force in the vertical direction. And it is easy to see that the normal force in the vertical direction is just N = m g- F sin , So the coefficient of friction is = f m g- F sin . 003 (part 1 of 2) 10.0 points Hint: This problem requires a train of logic. (1) Analyze force diagram, (2) use Newtons Laws, and (3) solve the equations of motion. A block starts from rest at a height of 7 . 7 m on a fixed inclined plane. The acceleration of gravity is 9 . 8 m / s 2 . 4 k g = . 1 2 27 What is the speed of the block at the bot- tom of the ramp? Correct answer: 10 . 7413 m / s. Explanation: Let : h = 7 . 7 m , m = 4 kg , = 0 . 12 , = 27 , and v f = final speed . reddy (ar38357) Friction clancy (SCI411-2) 2 F f N m g 27 d h 27 The normal force to the inclined plane is N = m g cos . The sum of the forces par- allel to the inclined plane is F net = m a = m g sin - m g cos a = g sin - g cos Since v 2 f = v 2 + 2 a x = 2 a d (1) along the plane and the distance moved along...
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