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# phyexam2 - Version 117 Midterm 2 orin(58140 This print-out...

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Version 117 – 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 54 . 7 m is banked so that a car traveling with uniform speed 54 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 . 3 Mg μ 0 θ What is θ ? 1. 28.5481 2. 18.928 3. 18.2 4. 22.7692 5. 21.6979 6. 29.7788 7. 34.0374 8. 26.1907 9. 32.3251 10. 31.1018 Correct answer: 22 . 7692 . Explanation: Let : m = 2300 kg , v = 54 km / hr , r = 54 . 7 m , and μ 0 . 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 m g 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 = m a 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 = m g (2) ( m g ) bardbl = m g sin θ (3) m a 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 = m a c = m v 2 r The component of this force parallel to the incline is summationdisplay vector F bardbl = m g sin θ = F net cos θ = m v 2 r cos θ .

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Version 117 – Midterm 2 – florin – (58140) 2 In this reference frame, the car is at rest, which means that the net force on the car (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 m g sin θ = F bardbl = m v 2 r cos θ tan θ = v 2 g r = (54 km / hr) 2 (9 . 8 m / s 2 ) (54 . 7 m) × parenleftbigg 1000 m km parenrightbigg 2 parenleftbigg hr 3600 s parenrightbigg 2 = 0 . 419729 θ = arctan(0 . 419729) = (0 . 397398 rad) bracketleftbigg 180 deg π rad bracketrightbigg = 22 . 7692 . 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 . What will be the object’s final speed if you do twice as much work? 1. v 2 2. 2 v 3. Still v 4. 4 v 5. 2 v correct Explanation: W = Δ K = 1 2 m v 2 v 2 so W v 2 and v 2 v 2 = 2 W W v = 2 v .
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