Version 001 Quest 3 Forces tubman (20131) 1


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1 Version 001 Quest 3 Forces tubman (20131) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. l B Conceptual 04 Q points In order to slide a heavy desk across the floor at constant speed in a straight line, you have to exert a horizontal force of 800 Newtons. Compare the 800Newton horizontal pushing force F to the frictional force f between the desk and the ground. F = f from Newton s third law. It depends on the direction the desk moves. F < f; the frictional force is greater than the pushing force since it is such a heavy desk. F = f from Newton s first law. correct F > f; the frictional force is always less than any other force acting on the object. d W p W l A Which way does the force point at position A? None of these s correct 7. Since the desk moves at constant speed in a straight line, Newton s first law tells us the two forces must balance. Conceptual forces (part 1 of 4) 10.0 points A ladder leans against a wall while someone climbs up, as shown in the figure below. The force points upward so that it doesn t fall through the floor. 003 (part 2 of 4) 10.0 points Which way does the force point at position B?
2 Version 001 Quest 3 Forces tubman (20131) 2 correct 7. None of these 7. If both the floor and the wall were frictionless, the base of the ladder would slide to the left. The force due to static friction points towards the right, opposing the ladder s tendency to move to the left. 005 (part 4 of 4) 10.0 points Whichwaydoestheforceduetofrictionpoint at position B? The wall has to push the ladder out so that it doesn t fall through the wall. The force thus points to the left. 004 (part 3 of 4) 10.0 points Considerthecasewhereboththewallandthe floor are rough. Which way does the force due to friction point at position A? correct None of these correct None of these graphs are correct. 7. If both the wall and the floor were frictionless, the top of the ladder would slide down.
3 Version 001 Quest 3 Forces tubman (20131) 3 The force due to static friction points upward, opposing the ladder s tendency to move downward. Friction on an Incline (part 1 of 3) 10.0 points Consider a block on an incline plane. The coefficients of static and kinetic friction between the block and the incline are µ s = 0.51 and µ k = 0.4 µ s = tanθ c θ c = tan 1 µ s = tan 1 (0.51) = (part 2 of 3) 10.0 points Determine the magnitude of the frictional force f when θ < θ critical. 1 kg µ θ c f = mgcosθ f = µ k mgcosθ f = mgsinθ correct Calculate the critical angle θ c, where the block just begins to slide. Correct answer: Basic Concepts: Friction: f s µ s N, f k = µ k N Let : m = 1 kg, µ k = 0.45, µ s = 0.51, and v f = final speed. N f = µ k mgsinθ f = µ s mgcosθ f = mgcotθ 7. f = µ s mgsinθ Below the critical angle the frictional force is exactly equal in magnitude and opposite in direction to the gravitational force parallel to the incline. From the force diagram we have f = mg = mgsinθ. mgsinθ mg f s θ c mgcosθ 008 (part 3 of 3) 10.0 points Determine the magnitude of the frictional force f when θ > θ critical. f = mgtanθ f = µ s mgsinθ Solution: The frictional force, f s, of a body at rest has to be less than µ s N = µ s mgcosθ. At θ = θ c, f s = µ s N. Since there is no acceleration along the plane of the incline, f s mgsinθ = 0. Thus, f s = µ s mgcosθ c = mgsinθ c f = mgcosθ f = mgcotθ f = µ k mgsinθ f = µ k mgcosθ correct
4 Version 001 Quest 3 Forces tubman (20131) 4 7. f = µ s mgcosθ When the block is sliding the frictional force is equal to the coefficient of sliding friction times the force. f = µ k F N = µ k mg = µ k mgcosθ. Sliding occurs when θ > θ critical. Hitting the Brakes 009 (part 1 of 3) 10.0 points You are driving at the speed of 25 m/s ( mph) when suddenly the car in front of you (previously traveling at the same speed) brakes. Considering an average human reaction, you press your brakes s later. Assume that the brakes on both cars are fully engaged and that the coefficient of friction is between both cars and the road. The acceleration of gravity is 8 m/s 2. Calculate the acceleration of the car in front of you when it brakes. Correct answer: 2218 m/s 2. The only force on the car horizontally when braking is the force of friction, F f. Thus F net = ma = F f = µn = µmg since the braking force is a deceleration. Thus d br = v2 0 2a = (25 m/s)2 2( 2218 m/s 2 ) = m 011 (part 3 of 3) 10.0 points Find the minimum safe distance at which you can follow the car in front of you and avoid hitting it (in the case of emergency braking described here). Correct answer: 1189 m. The car in front of you moves a total distance of d br, which also is the distance you travel while braking. You also traveled a distance d react = v 0 t because of your reaction time t. Thus the minimum safe distance is the distance you travel during your reaction time: d safe = v 0 t = (25m/s)(0.478s) = 1189m Conceptual forces 06 short points A spherical mass rests upon two wedges, as seen in the figure below. The sphere and the wedgesareatrestandstayatrest. Thereisno friction between the sphere and the wedges. M ma = µmg a = µg = (0.941)(8m/s 2 ) = 2218m/s (part 2 of 3) 10.0 points Calculate the braking distance for the car in front of you. Correct answer: m. For constant acceleration a, v 2 v 2 0 = 2a(x x 0). Thefinalvelocityv f = 0,andford br = x x 0, The following figures show several attempts at drawing freebody diagrams for the sphere. Which figure has the correct directions for each force? Note: The magnitude of the forces are not necessarily drawn to scale. v 2 0 = 2ad br
5 Version 001 Quest 3 Forces tubman (20131) 5 correct Weight the force of gravity pulls the sphere down. The force of the left wedge upon the sphere acts perpendicular to ( to) their surfaces at the point of contact; i.e., diagonally upward and rightward. Likewise, the force of the right wedge upon the sphere acts diagonally upward and leftward. Free Body Diagram of Balloon 013 (part 1 of 2) 10.0 points A balloon is waiting to take off. As seen in the figure below, the balloon s basket sits on a platform which rests on the ground. The balloon is pulling up on the basket, but not hard enough to lift it off the platform. balloon platform basket ground What is the freebody diagram for the platform? Some parallel vectors are offset horizontally for clarity. friction friction F ground on plat F bask on plat correct 7. Since the sphere is not moving, no forces act on it. F bal on plat F plat on bask F plat on ground
6 Version 001 Quest 3 Forces tubman (20131) 6 F bal on plat F plat on ground 0.99 m on a horizontal, frictionless table. The other end of the string passes through a hole inthecenterofthetableandamassof0.72kg is tied to it. The suspended mass remains in equilibrium while the puck revolves. All forces acting on the platform must be represented. No forces acting on other objects should be represented. v 0.99 m kg 014 (part 2 of 2) 10.0 points What is the free body diagram for the basket? Again some parallel vectors are offset horizontally for clarity kg What is the tension in the string? The acceleration due to gravity is 8 m/s 2. F plat on bask correct Correct answer: N. F bask on plat F bask on bal Let : M = 0.72 kg and g = 8 m/s 2. F ground on bask v r m F bask on plat F bask on bal All forces acting on the basket must be represented. No forces acting on other objects should be represented. Serway CP (part 1 of 3) 10.0 points Anairpuckofmass0.315kgistiedtoastring and allowed to revolve in a circle of radius Mg Since the suspended mass is in equilibrium, the tension is T = M g = (0.72 kg) ( 8 m/s 2) = N. 016 (part 2 of 3) 10.0 points What is the horizontal force acting on the puck? Correct answer: N. The horizontal force acting on the puck is the tension in the string, so F c = T = N.
7 Version 001 Quest 3 Forces tubman (20131) (part 3 of 3) 10.0 points What is the speed of the puck? Correct answer: m/s. At the poles, the rotating velocity is zero, so F c = W a mg = 0 W a = mg = (137 kg) ( 8 m/s 2) = N. F c = mv2 r v = Fc r m = = m/s. (7.056 N)(0.99 m) kg Serway CP (part 1 of 2) 10.0 points Because of Earth s rotation about its axis, a point on the Equator experiences a centripetal acceleration of m/s 2, while a point at the poles experiences no centripetal acceleration. What istheapparent atthe equator of a person with a mass of 137 kg? The acceleration of gravity is 8 m/s 2. Your answer must be within ± 0.1% Correct answer: N. Let : a c = m/s 2 and m = 137 kg. Since the centripetal acceleration of a person is inward (toward the Earth s axis), F c = ma c = mg W a W a = mg ma c = m(g a c ) = (137 kg) ( 8 m/s m/s 2) = N. 019 (part 2 of 2) 10.0 points What is his apparent at the poles? Correct answer: N.
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