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# ACADEMIC CONTIN EXPER FOR SUCC ACAF 100

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This 6 page Class Notes was uploaded by Carlie Watsica on Monday September 28, 2015. The Class Notes belongs to ACAF 100 at Old Dominion University taught by Lisa Hall in Fall. Since its upload, it has received 33 views. For similar materials see /class/215323/acaf-100-old-dominion-university in Academic Contin Exper at Old Dominion University.

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Date Created: 09/28/15

Review For Midterm Exam 1 Problem 1 vectors Given vectors AT 572179 60 and BT 472179 730 7 nd the magni tude and the angle of the vector 739 fl Problem 2 vectors A vector C has magnitude of 85m and makes and angle of 7215 i lf 739 11 and AT 572179 105 7 nd vector Bi Problem 3 1D motion A car travels 10 sec with acceleration a 2 721327 then travels 15 min with a constant speed Then reduces the speed and travels 10 km in the next 20 mini Find a the velocity of the car after the rst 103657 b the total distance traveled7 c the velocity of the car in the 3rd segment and d the average velocity of the trip Problem 4 2D motion A rock is thrown from the top of the 60 m high building After 14 sec it hits the building 140 m away at height 20 mi Find a vac and vy b the angle of launch7 c highest point of trajectory7 d the time that it takes to reach the highest point7 e the angle of velocity just before it hits the building Problem 5 Newton s Laws Two masses with m1 3 kg and 7212 5 kg are connected with a string and laying on the table top and a mass m3 5 kg is is hangin from the side of the table connected to the mass 7212 through the string passing through the pulley Coef cients of kinetic frictions for masses m1 and 7212 are 1 0 1 and M2 006 respectively Find a acceleration of the system7 b the tension of the string connecting m1 and 7212 and c the tension of the string connecting 7213 an mgi Problem 6 Newton s Laws The block of mass m Sky is sliding on the incline plane of 30 with a 6 Til82 Find the coef cient of kinetic frictioni Problem 7 kinematics and Newton s laws A box sliding With initial velocity v 13 ms comes to a complete stop after 20 m7 Find the coef cent of kinetic frictioni SOLUTIONS Solution problem 1 First we need to decompose the vectors fl and A75 lAl cos60O 25m 7 Ag lAl sin60 43m B75 lBl cos 330 35m 7 By lBl sin330 72m Note 730 corresponds to 330 measured counter clockwise from positive direction of Xaxisi The resulting vector C Will be C75 Ag Bag 2i5m 35m 6m Cy Ag By 4i3m7 2m 23m 0 03 05 M36 529 64m 90 arctg 21 1 Solution problem2 First step is to decompose given vectors into the components A75 lAl cos 105 713721 Ag lAl sin105o 48m C75 lCl cos 145 77m Cy lCl sinl45 49m Now7 from equations CmAgcBgc7 CyAyBy Weget BECxiAE 7 777217 713721 757721 By Cy 7 Ag 49m 7 48m Oilm So the magnitude of the vector E B B B x3249 001 57 B 9 arctgF 17890 Solution problem 3 a First we must nd the velocity after the rst segment we have vf v0at 0ms2 72132 10 86620 ms And the distance covered by the car in the rst 10 sec will be 1 z 710 v0tat22 0 721er E2 ms 10 sec2 200 m 02 km In order to calculate distances in km we need to transform the velocity ms A cmhour Since there are 1000 m in kilometer and 3600 sec in an hour 7 vm1000mkm 7 lsec3600sech0ur b for total distance we need the distance covered in the 2nd segment which the car traveled with constant speed of 72 kmh solution to all 72 kmhour d2 vt 72 kmh 025 hour 18 km now if we add all the distances we get D d1d2d3 02 km18 km10 km 282 km C the car traveled 10 km in 20 min which means 7 g 7 10 km v3 7 t 7 033 hour cl for average speed we just need the total distance and the total time 303 cmhquot 7 dmal 39Uav 7 tuna Total distance is dtotal d1 d2 d3 282 km Total Time is tuna 751 752 753 0003 hquot 025 hquot 033 hquot 058 hquot and the average velocity is 7 282 km 7 7 058 m 486 cmhquot Solution 4 problem 4 We set up our axis so the y 0m corresponds to ground level a With the given information the vac is just 7 d7 7 140 m v1 7 7 7 14 sec where dag is the horizontal distance covered To nd vy we need to use the equation for yaxis only 10 ms 1 2 yyovot7 iyt where y is the nal coordinate of the projectile and yo is the initial We know the time and the gravity constant 1 20 m 60 721 pay14 s 7 E98 m32l4 s2 740mv0y1437960 4m v0y l4 s voy 657 ms b The angle of launch will be voy 657 0 7 7 813 um 10 c to nd the highest point we rst nd the time it takes for the vy 0 ms7 using a arctg 1 657 ms vyv0yigt tm67sec and the y coordinate after 67 sec will be 1 2 yy0v0yt gyt y 60 721 65 ms 67 3 7 98 ms2 67 s2 60 721 220 m 280 m d The time we already found to be t1 67 sec e In order to nd the angle at the impact we need to know the velocity components of the projectile at the impact time 14 sec after launch we use vfv0at for each of the velocity components 1ng vw0m32l4s10 ms vfy voy igt 65 7213798 m3214 s 7715 s The direction of the velocity in the y direction is opposite to the launch velocity The impact angle Will be a arctg 771 5 mS 782 1ng 10 ms Solution problem 5 In the solution the gravity constant 9 10 72132 is used Find the acceleration of the system a The force acting on blocks m1 and 7212 in the positive direction is the gravitational force of mg equal to Fmgg5kglOm3250N Forces opposing the motion are the friction forces on m1 and 7212 equal to f1M1m190i13kg10m323N f2 Mgmgg0i065 Icy10 72132 SN So7 the acceleration Will be Fm F f1 f2 50N73N73N a7 m1m2 m1m2ms 3kg5kg5ky MN 2 a7 8kg 73i4ms b The mass m1 is being accelerated With a rate of 34 72132 and the forces acting on it are only the friction force and the tension forse Tl7 the equation of second laW of Newton reads T1 f1m10 T1m1af1 T13kg3i4ms23N10i2N3N13i2N c To nd the tension force on 7212 we again consider all the forces acting on mgr They are Tension force T1 from the block m1 7 friction force 7 and the tension force in the string connecting the mg to mgr So the Newton7s 2nd laW reads TQ T1 f2m2a T2m2aT1f2 T25kg3l4ms213l2N3N33l2N Solution problem 6 We rst need to decompose the gravity force into I and y componentsl We direct our axis 1 to be parallel to the incline plane and y perpendicular The positive z direction will be down the plane Again 9 10 72132 is used In this setup the 1 component of the gravity force will be W75 Wcos 30 5 kg 10 ms2003300 433 N Wy Wsin30 5 kg 10 ms2sm30 25 N Since there is no acceleration in the y direction the Wy should be balanced with a normal force And the friction force is proportional to the normal force equal in magnitude to Wyl f Wy Now forces driving the box in the z direction are the Was in positive z direction and f in negative W75 7 f ma Since we know the mass and the acceleration of the object we can nd the friction force from the equation above then knowing the normal force calculate the coef cient of static frictionl fW 7ma a 433N75kg6ms213l3N 133 N 133 N W 133 N f M y f M Wy 25 N In reality the mass of the box is not neccesary to be given it cancels out in the nal equations 0 53 01 Solution problem 7 In order to nd the friction coef cient we need to nd the acceleration of the box decceleration in our case For that we use 112 113 2az 7 ID

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