This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 086 Midterm 2 florin (58140) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A curve of radius 52 . 9 m is banked so that a car traveling with uniform speed 47 km / hr can round the curve without relying on fric tion to keep it from slipping to its left or right. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 9 M g What is ? 1. 17.8004 2. 29.4517 3. 21.7496 4. 33.6484 5. 28.5481 6. 31.2492 7. 32.3251 8. 22.4695 9. 27.0 10. 18.2 Correct answer: 18 . 2 . Explanation: Let : m = 2900 kg , v = 47 km / hr , r = 52 . 9 m , and . Basic Concepts: Consider the free body diagram for the car. The forces acting on the car are the normal force, the force due to gravity, and possibly friction. N N N cos mg N sin x y To keep an object moving in a circle re quires a force directed toward the center of the circle; the magnitude of the force is F c = ma c = m v 2 r . Also remember, vector F = summationdisplay i vector F i . Using the freebody diagram, we have summationdisplay i F x N sin N cos = m v 2 r (1) summationdisplay i F y N cos + N sin = mg (2) ( mg ) bardbl = mg sin (3) ma bardbl = m v 2 r cos (4) and , if = 0 , we have tan = v 2 g r (5) Solution: Solution in an Inertial Frame: Watching from the Point of View of Some one Standing on the Ground. The car is performing circular motion with a constant speed, so its acceleration is just the centripetal acceleration, a c = v 2 r . The net force on the car is F net = ma c = m v 2 r The component of this force parallel to the incline is summationdisplay vector F bardbl = mg sin = F net cos = m v 2 r cos . In this reference frame, the car is at rest, which means that the net force on the car Version 086 Midterm 2 florin (58140) 2 (taking in consideration the centrifugal force) is zero. Thus the component of the net real force parallel to the incline is equal to the component of the centrifugal force along that direction. Now, the magnitude of the cen trifugal force is equal to F c = m v 2 r , so F bardbl = F net cos = F c cos = m v 2 r cos F bardbl is the component of the weight of the car parallel to the incline. Thus mg sin = F bardbl = m v 2 r cos tan = v 2 g r = (47 km / hr) 2 (9 . 8 m / s 2 ) (52 . 9 m) parenleftbigg 1000 m km parenrightbigg 2 parenleftbigg hr 3600 s parenrightbigg 2 = 0 . 328783 = arctan(0 . 328783) = (0 . 317649 rad) bracketleftbigg 180 deg rad bracketrightbigg = 18 . 2 . 002 10.0 points You do a certain amount of work on an object initially at rest, and all the work goes into increasing the objects speed. If you do work W , suppose the objects final speed is v ....
View
Full
Document
 Spring '10
 Florin

Click to edit the document details