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Unformatted text preview: Version 086 – Midterm 2 – florin – (58140) 1 This print-out should have 25 questions. Multiple-choice 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 μ ≈ θ What is θ ? 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 μ ≈ . 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. μ N N N cos θ mg N sin θ 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 free-body diagram, we have summationdisplay i F x N sin θ − μ N cos θ = m v 2 r (1) summationdisplay i F y N cos θ + μ N sin θ = mg (2) ( mg ) bardbl = mg sin θ (3) ma bardbl = m v 2 r cos θ (4) and , if μ = 0 , we have tan θ = 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 θ = F net cos θ = m v 2 r cos θ . 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 θ = F c cos θ = m v 2 r cos θ F bardbl is the component of the weight of the car parallel to the incline. Thus mg sin θ = F bardbl = m v 2 r cos θ tan θ = 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 θ = arctan(0 . 328783) = (0 . 317649 rad) bracketleftbigg 180 deg π 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 object’s speed. If you do work W , suppose the object’s final speed is v ....
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