Version 001 test 1 review tubman (IBII201516) 1


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1 Version 001 test 1 review tuban (IBII01516) 1 This printout should have 44 questions. Multiplechoice questions ay continue on the next colun or page find all choices before answering. Crossbow Experient 001 (part 1 of 3) 10.0 points In deep space (no gravity), the bolt (arrow) of a crossbow accelerates at 118 /s and attains a speed of 11 /s when it leaves the bow. For how long is it accelerated? Correct answer: s. Let : a 118 /s and v 11 /s. v v o +at at t v a 11 /s 118 /s s. 00 (part of 3) 10.0 points What speed will the bolt have attained.9 s after leaving the crossbow? Correct answer: 11 /s. It has left the crossbow, so a 0 and the velocity of the bolt after.9 s will reain 11 /s. 003 (part 3 of 3) 10.0 points How far will the bolt have traveled during the.9 s? Correct answer: s s o +v o t+ 1 at vt (11 /s)(.9 s) Holt SF 0D (part 1 of ) 10.0 points A car with an initial speed of.8 k/h acceleratesatauniforrateof0.95/s for4. s. Find the final speed of the car. Correct answer: /s. Let : v i.8 k/h, a 0.95 /s, and t 4. s. v f v i +a t (.8 k/h) k 1 h 3600 s +(0.95 /s )(4. s) /s. 005 (part of ) 10.0 points Find the displaceent of the car after that tie. Correct answer: k. x v i t+ 1 a( t) 1 h (.8 k/h)(4. s) 3600 s + 1 ( 0.95 /s ) (4. s) 1 k k. Particle Motion points The plot shows x(t) for a particle. x t
2 Version 001 test 1 review tuban (IBII01516) Which stateent is correct about the otion? 1. The particle at first speeds up, then slows down, then speeds up again.. The particle follows the path of a projectile, like a thrown baseball. 3. The particle oves along a portion of a circle, as the sketch shows directly. 4. The particle oves at constant speed, but not at constant velocity. 5. The particle slows down, coes oentarily to a rest, then returns with increasing speed to its starting point. correct Since v x is the slope of x(t), the particle slows down, coes to oentary rest, and then reverses direction, increasing speed as it returns to its starting point. Velocity vs Tie (part 1 of 8) 10.0 points Consider the plot below describing otion along a straight line with an initial position of velocity (/s) tie (s) What is the acceleration at 1 second? Correct answer: 3.5 /s. The slope of the velocity curve fro 0 seconds to seconds is a v f v i t f t i 3.5 /s. 7 /s 0 /s s 0 s 008 (part of 8) 10.0 points What is the velocity at seconds? Correct answer: 7 /s. The velocity at seconds can be read fro the plot; however, it can also be calculated: v v i +a(t f t i ) 0 /s+(3.5 /s )( s 0 s) 7 /s. 009 (part 3 of 8) 10.0 points What is the position at seconds? Correct answer: 17. velocity (/s) Let : x tie (s)
3 Version 001 test 1 review tuban (IBII01516) 3 Thepositionatsecondsisx 0 plusthearea of the triangle (shown in gray) x ( s 0 s)(7 /s 0 /s) 17 ; however, it can also be calculated: x x i +v i (t f t i )+ 1 a(t f t i ) 10 +(0 /s)( s 0 s) (3.5 /s )( s 0 s) 010 (part 4 of 8) 10.0 points What is the acceleration at 4 seconds? Correct answer: 1 /s. The slope of the velocity curve fro seconds to 6 seconds is a v f v i t f t i 1 /s. 3 /s 7 /s 6 s s 011 (part 5 of 8) 10.0 points What is the position at 6 seconds? Correct answer: 37. The position is 17 plus the area of the trapezoid x (6 s s)(3 /s+7 /s) 37 ; however it can also be calculated: x x i +v i (t f t i )+ 1 a(t f t i ) 17 +(7 /s)(6 s s) + 1 ( 1 /s )(6 s s) (part 6 of 8) 10.0 points What is the acceleration at 8 seconds? Correct answer: /s. The slope of the velocity curve fro 6 seconds to 9 seconds is a v f v i t f t i /s. 4 /s 0 /s 9 s 6 s 013 (part 7 of 8) 10.0 points What is the velocity at 8 seconds? Correct answer: /s. The velocity at 8 seconds can be read fro the plot; however, it can also be calculated: v v i +a(t f t i ) 0 /s+( /s )(8 s 6 s) /s. 014 (part 8 of 8) 10.0 points What is the position at 8 seconds? Correct answer: The position is 37 plus the area of the triangle x (8 s 6 s) ( /s 0 /s) ; however it can also be calculated: x x i +v i (t f t i )+ 1 a(t f t i ) 37 +(0 /s)(8 s 6 s) + 1 ( /s )(8 s 6 s)
4 Version 001 test 1 review tuban (IBII01516) 4 Spider in a Web points A spider of ass is sitting in a perfectly syetric web (in a vertical plane with 45 angles between the web lines) as shown in the figure below, where T i s denote the tensions alongthelineswhichareconnectedtothespider at the center with i 1,,3,4,5,6,7,8. The tension are all positive. T 5 T 4 T 3 T T 6 T 7 T 8 How is the tension T 3 related to T, T 8, and T 7? Hint: Fro leftright syetry, we see that T 1 T 5, T T 4, and T 8 T T 3 T 7 (T 8 T ) g. T 3 T 7 (T 8 +T ) g 3. T 3 T 7 4. T 3 T 7 +(T 8 +T ) +g 5. T 3 T 7 +g 6. T 3 T 7 +(T 8 +T ) g T 1 7. T 3 T 7 +(T 8 T ) +g correct 8. T 3 T 7 (T 8 T ) +g 9. T 3 T 7 (T 8 +T ) +g 10. T 3 T 7 +(T 8 T ) g Basic Concept Newton s nd Law Solution: In the x (horizontal) direction, we have (T +T 8 ) 1 +T 1 (T 4 +T 6 ) 1 T 5 0, so, T 1 T 5, T T 4, and T 8 T 6. In the y (vertical) direction, we have 1 (T +T 4 ) +T 3 (T 8 +T 6 ) 1 T 7 g 0, andusing theconditiont T 4 andt 8 T 6, T +T3 T 8 T7 g 0, or T 3 T 7 +(T 8 T ) +g. Weather Balloon Acceleration 016 (part 1 of 4) 10.0 points The instruents attached to a weather balloon have a ass of 4.3 kg. The balloon is released and exerts an upward force of 98.7 N on the instruents. The acceleration of gravity is 9.8 /s. What is the acceleration of the balloon and the instruents? Correct answer: /s. Basic Concepts: F a. Solution: The balloon exerts an upward force F upward and gravity acts down, so F net a net F g a net F g directed upward N (4.3 kg)(9.8 /s ) 4.3 kg /s. 017 (part of 4) 10.0 points After the balloon has accelerated for 1.7 s, the instruents are released.
5 Version 001 test 1 review tuban (IBII01516) 5 What is the velocity of the instruents at the oent of their release? Correct answer: /s. The instruents are traveling the sae speed as the balloon at the tie of their release. The net acceleration is upward and v o 0, so directed upward. v v o +a net t ( /s )(1.7 s) /s. 018 (part 3 of 4) 10.0 points Assue: The upward direction is positive. What net force acts on the instruents after their release? Correct answer: 4.14 N. Once released fro the balloon, the only acceleration is that due to gravity, so F net g (4.3 kg)(9.8 /s ) 4.14 N. 019 (part 4 of 4) 10.0 points How long after their release does the direction of their velocity first becoe downward (negative)? Correct answer: s. Upon release, their initial upward velocity is v and gravity acts downward. They start oving downward when the velocity is 0, so v f v o gt 0 v gt t v g /s 9.8 /s s. Person in an Elevator points Consider a person standing in an elevator that is accelerating upward. The upward noral force N exerted by the elevator floor on the person is 1. larger than the weight (gravitational force) of the person. correct. saller than the weight (gravitational force) of the person. 3. identical to the weight (gravitational force) of the person. According to Newton s nd Law, the equation of otion for the person on the elevator is N W a, so N W +a > W. Horse and Cart points Iagine a horse pulling a cart; the horse and the cart ove at the sae velocity. Whether the horse and the cart are oving at a constant velocity or not is not iportant. Now, accordingtothelawsofotion,thehorseand the cart exert forces on each other. Which force is bigger? 1. Force exerted by the cart on the horse. Insufficient inforation to deterine 3. Force exerted by the horse on the cart 4. Two forces are equal correct Newton s 3 rd law can be applied here to explain that the two forces are equal.
6 Version 001 test 1 review tuban (IBII01516) 6 AP M 1998 MC 1 0 (part 1 of ) 10.0 points A block of ass is accelerated across a rough surface by a force of agnitude F that is exerted at an angle φ with the horizontal, as shown above. The frictional force on the blockexertedbythesurfacehasagnitudef. f What is the agnitude of the acceleration a of the block? 1. a F cosφ. a F cosφ f correct 3. a F sinφ g 4. a F f 5. a F Fro Newton s second law of otion, the acceleration is the total force in the horizontal direction divided by the ass. There are two forces in the horizontal direction: one is the friction force; the other is the horizontal coponent of the dragging force F, but they are in the opposite directions, so the acceleration of the block is a F cosφ f 03 (part of ) 10.0 points Which of the following expressions for the coefficient of friction is correct? 1. µ g f f. µ g F cosφ F. φ 3. µ g F cosφ f 4. µ f g f 5. µ g F sinφ correct By definition, the coefficient of kinetic friction is the ratio of the friction force and the noral force in the vertical direction. And it is easy to see that the noral force in the vertical direction is just N g F sinφ, So the coefficient of friction is µ f g F sinφ. Accelerated Blocks points Twoblocksarearrangedattheendsofaassless string as shown in the figure. The syste starts fro rest. When the 3.6 kg ass has fallen through 0.415, its downward speed is 1.33 /s. The acceleration of gravity is 9.8 /s. a 4.8 kg µ 3.6 kg What is the frictional force between the 4.8 kg ass and the table? Correct answer: N. Given : kg, 4.8 kg, v 0 0 /s, and v 1.33 /s.
7 Version 001 test 1 review tuban (IBII01516) 7 Basic Concept: Newton s Second Law F M a Solution: Theaccelerationof 1 isobtained fro the equation v v 0 a(s s 0 ) a v v 0 h (1.33 /s) (0 /s) (0.415 ).131 /s. Consider free body diagras for the two asses µn N a g T T a 1 g Because 1 and are tied together with string,theyhavesaethespeedandthesae acceleration, so the net force exerted on is F a The net force on 1 is 1 a 1 g T, so that T 1 g 1 a. Thus F T f k, f k T F 1 g ( 1 + )a (3.6 kg)(9.8 /s ) (3.6 kg+4.8 kg) (.131 /s ) N. Haer Throw 05 (part 1 of ) 10.0 points An athlete whirls a 7.8 kg haer tied to theendofa1.5chaininasiplehorizontal circle where you should ignore any vertical deviations. The haer oves at the rate of rev/s. What is the centripetal acceleration of the haer? Assue his ar length is included in the length given for the chain. Correct answer: /s. Convert revolutions per second to radians per second. π(0.734 rev/s) rad/s. The centripetal acceleration is a c v r (rω) rω r (1.5 )( rad/s) /s. 06 (part of ) 10.0 points What is the tension in the chain? Correct answer: N. The chain supplies the centripetal force, so its tension is T a c (7.8 kg)( /s ) N. SWCT Centrifugal Force points You are a passenger in a car and not wearing your seat belt. Without increasing or decreasing its speed, the car akes a sharp left turn, and you find yourself colliding with the righthand door. Which is the correct analysis ofthe situation? 1. Neither of these. Before and after the collision, there is rightward force pushing you into the door. 3. Starting at the tie of collision, the door exerts a leftward force on you. correct
8 Version 001 test 1 review tuban (IBII01516) 8 4. Both of these Your otion reains in a straight line by Newton s first law. To keep you in the car that is turning left the door exerts a leftward force on you. Satellite points A counication satellite does not fall to the earth A. only if it is in a geosynchronous orbit. B. because the net force on it is zero. C. because it is beyond the pull of the earth s gravity. D. because it is in the earth s gravitational field. E. because it is being pulled by the sun and by other planets as well as the earth. 1. C only. D only 3. B only 4. E only 5. B and C 6. A only 7. None of these correct The satellite does not fall to the earth because it is in a unifor circular otion where the net force on it, ainly the Earth s gravity, equal to the centripetal force needed for the circular otion. Final Speed points You do a certain aount 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 uch work? 1. v correct. v 3. 4v 4. v 5. Still v W K 1 v v so v v W W v v. W v and Kinetic Energy of a Car points A(n) 1554 kg car travels at a speed of 11.8 /s. What is its kinetic energy? Correct answer: J. Kinetic energy is K 1 v 1 (1554 kg)(11.8 /s) J Block on an Inclined Plane (part 1 of ) 10.0 points A block of ass is pushed a distance D up aninclinedplanebyahorizontalforcef. The plane is inclined at an angle θ with respect to the horizontal. The block starts fro rest and the coefficient of kinetic friction is µ k.
9 Version 001 test 1 review tuban (IBII01516) 9 F D IfN isthenoralforce; i.e.,thesu ofthe Fcoponent which is perpendicular to the inclined plane and the gcoponent which is perpendicular to the inclined plane, what is the work done by friction? µ k 1. W +µ k (N gcosθ)d. W µ k (N gcosθ)d 3. W +µ k N D 4. W +µ k (N +gcosθ)d θ correct 6. v (F cosθ µ kn)d 7. v (F cosθ +gsinθ µ kn)d 8. v (F sinθ µ kn)d The work done by gravity is W grav gdcos(90 +θ) gdsinθ The work done by the force F is W F F Dcosθ. Fro the workenergy theore we know that W net K, 5. W 0 6. W µ k N D correct 7. W µ k (N +gcosθ)d The force of friction has a agnitude F friction µ k N. Since it is in the direction opposite to the otion, we get W friction F friction D µ k N D. 03 (part of ) 10.0 points What is the final speed of the block? 1. v (F cosθ gsinθ+µ kn)d. v (F sinθ+µ kn)d 3. v (F cosθ gsinθ)d 4. v (F cosθ+gsinθ)d 5. v (F cosθ gsinθ µ kn)d W F +W grav +W friction 1 v f. Using the expressions for the works, we get v f (F cosθ gsinθ µ kn)d. Coent: One ay work this proble in a systeatic way. Begin with the workenergy theore: W ext i >f K f K i +U f U i +W dissip i >f W ext i >f F cosθd K f K i 1 v f U f U i gd sinθ W dissip i >f µn D. This leads to the sae answer as given above. Work Done by Friction points A block sliding on a horizontal surface has an initial speed of 0.5 /s. The block travels a distance of 1 as it slows to a stop.
10 Version 001 test 1 review tuban (IBII01516) 10 What distance would the block have traveled if its initial speed had been 1 /s? correct 6. ore inforation is needed to answer the question W nc f k L 0 1 v 0 µgl 1 v 0 L v 0 µg So the distance is proportional to the square of the initial speed. Frictional Force on a Car points When an autoobile oves with constant velocity, the power developed is used to overcoe the frictional forces exerted by the air and the road. If the engine develops 8 hp, what total frictional force acts on the car at 7 ph? One horsepower equals 746 W, and one ile is Correct answer: N. F P v 8 hp 746 W hp 7 ph 1609 h i 3600 s N Pushing a Wheelbarrow (part 1 of ) 10.0 points Robin pushes a wheelbarrow by exerting a 141Nforce horizontally. Robinovesit87 at a constant speed for 4 s. What power does Robin develop? Correct answer: W. Let : F 141 N, d 87, and t 4 s. Since W Fd, the power becoes P W F d t t (141 N)(87 ) W. 4 s 036 (part of ) 10.0 points If Robin oves the wheelbarrow 6 ties as fast, how uch power is developed? Correct answer: W. Let : n 6. Since P F d F d t t F v, P v for a constant force, and P new P v new v nv v n Apply P E t Fd t Fv P new np 6( W) W. Pushing a Cart 037 (part 1 of ) 10.0 points Consider pushing a cart up each of the four
11 Version 001 test 1 review tuban (IBII01516) 11 frictionless raps. The cart begins at rest to the left of each rap and then ends at rest at the top. Let W A, W B, W C and W D be the aount of work needed to push the cart up each rap. A B C D What describes the relationship between the work required in each case? 1. W A < W B < W C < W D. W A W B W C W D 3. W A < W B W C W D 4. W A W B W C < W D 5. W A > W B > W C > W D 6. W A > W B W C W D correct The work done to the block is equal to the change of the potential energy, gh, with h the height of the rap, since the cart is at rest at both the botto and the top. According to the figure, B, C and D have the sae height h and only A has a larger h, the relationship between the work done should be W A > W B W C W D. 038 (part of ) 10.0 points Justaseachcartreachesthetopofeachrap, it is released and rolls back down to the left. Justasitreachesthefloor,itsvelocityoneach rap is v A, v B, v C and v D. Which of the following describes the relationship between the final velocities in each case? 1. v A > v B > v C > v D. v A v B v C < v D 3. v A < v B < v C < v D 4. v A v B v C v D 5. v A < v B v C v D 6. v A > v B v C v D correct After the cart is released and rolls down to the botto, the potential energy changes into kinetic energy. Only A has a higher potential, so its speed at the botto is also larger than the others while B, C and D have the sae speed at the botto. Rolling Down a Driveway points A 685 kg car starts fro rest at the top of a driveway of length 4 that is sloped at 4 with the horizontal. An average frictional force of 380 N ipedes the otion. Find the speed of the car at the botto of the driveway. The acceleration of gravity is 9.8 /s. Correct answer: /s. The initial vertical height of the car above the zero reference level at the base of the hill is y i l sinθ 4 sin The energy lost through friction is We now use W f F l (380 N)(4 ) 950 J. E k0 +E p0 +W f E kf +E pf E p0 +W f 1 v (Ep0 +W f ) v [ J+( 950 J)] 685 kg /s
12 Version 001 test 1 review tuban (IBII01516) 1 Quarterback and Receiver 040 (part 1 of ) 10.0 points A(n) kg football is thrown with a speed of 9.75 /s. A stationary receiver catches the ball and brings it to rest in s. What is the agnitude of the ipulse delivered to the ball? Correct answer: kg /s. Let : kg and v i 9.75 /s. The ipulse delivered to the ball is F t p v f v i 0 (0.566 kg)(9.75 /s) kg /s, so the ipulse delivered to the ball is kg /s. 041 (part of ) 10.0 points What is the average force exerted on the receiver? Correct answer: N. F p t Let : t s kg /s s N. Students on Skates points Two students on roller skates stand facetoface, then push each other away. Which student will always have the fastest speed? 1. the student with the saller ass correct. Unable to deterine 3. the student with the larger ass Moentu is conserved, so v 1 + v 1 v 1 v v 1 v 1 The inus sign is ignored for speeds, so v 1 > v eans that > 1. Your Body in a Collision 043 (part 1 of ) 10.0 points An ipulse of 151 Ns is required to stop a person s head in a car collision. If the face is in contact with the steering wheel for s, what is the average force on the cheekbone? Correct answer: N. p 151 Ns and t s. Force is related to ipulse by F p t I p F t, so 151 Ns s N. 044 (part of ) 10.0 points If an average force of 905 N fractures the cheekbone, how long ust it be in contact with the steering wheel in order to fracture? Correct answer: s. t p F 1 Let : F N. 151 Ns 905 N s.
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