Final Exam Richland College 1. In the picture θ=25.0°, mass of the object is 7 kg and coefficient of kinetic friction is 0.19. Find the acceleration of the object when moves downward. (15 points) 2. Find the resultant of the three vectors in the picture. (20 points) 3. A student applied 6.0 J work to stretch a spring 2 cm from its relaxed position. How much work her friend should apply to stretch the spring 4 cm in addition. (10 points) 4. James had an accident when driving at the speed of 95 km/h. The car travels 0.8 m after hitting an object and before coming to a complete stop. Find the average acceleration of the car. (15 points) 5. In the picture, the mass of dart is 0.100 kg and the constant of the spring is 250 N/m (spring has negligible mass). The spring is compressed 6cm (First picture). What would be the velocity of the dart in the second picture when it leaves the gun while the spring is at equilibrium. (15 points) 6. A 10 kg box rests on a horizontal floor. Find the force of friction when the applied force is 20 N, 38 N, 40 N. The coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.3 (15 points) 7. Find the displacement from 1.5 seconds to 3.1 seconds if v(t)=25+18t (v is in m/s and t is in seconds) (10 points) Final Exam Richland College 1. In the picture θ=25.0°, mass of the object is 7 kg and coefficient of kinetic friction is 0.19. Find the acceleration of the object when moves downward. (15 points) 2. Find the resultant of the three vectors in the picture. (20 points) 3. A student applied 6.0 J work to stretch a spring 2 cm from its relaxed position. How much work her friend should apply to stretch the spring 4 cm in addition. (10 points) 4. James had an accident when driving at the speed of 95 km/h. The car travels 0.8 m after hitting an object and before coming to a complete stop. Find the average acceleration of the car. (15 points) 5. In the picture, the mass of dart is 0.100 kg and the constant of the spring is 250 N/m (spring has negligible mass). The spring is compressed 6cm (First picture). What would be the velocity of the dart in the second picture when it leaves the gun while the spring is at equilibrium. (15 points) 6. A 10 kg box rests on a horizontal floor. Find the force of friction when the applied force is 20 N, 38 N, 40 N. The coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.3 (15 points) 7. Find the displacement from 1.5 seconds to 3.1 seconds if v(t)=25+18t (v is in m/s and t is in seconds) (10 points) Formula Sheet . - Final One-dimensionalkinematics r displacemqfi: Ax = xI - xi . averagesp"*: totaldistancetraveled total time instantanous speed =fnf (mafritude of the instantaneous velocity) xf - xi averagevelocity: u* =+ = o . Lt tI -t, o instantaneous . av€rage acceleration: o instantaneous accel leration: o o velocity ,= li:or*=;Trf aou: -vi at =vr tf-ti L! = ,yoo*=ITr# One-dimmsional motion with constant acceleration (4 facts): (l) v =vo+ at (2) x=xo+)00+u)t (3) x=xo+vst+Lat2 (4) o Free 'r ,2 =ri, *Zo(*-ro) fall (positive direction fory taken tobe upword) x 1! and a -r -g in the above 4 facts: (1) v=vo-gt (z) y=yo+)Qo+u)t (3) y= yo+vot-)stz (4) u'=u3-zsb-Y) r Vectors The resultant vector fot several vectors is the vector sum. For examplg if you have three displacements the resultant displacerirent ,E is given by: , and ir, i, ir, fr.=ir+ir+ir. Ifa vector 7 o is unitten in component"for- as tr = of Getting magnitude and directim ll=,[e?T] .(,c \ 0 tan-tl = (magnitudeof 7 ) ? t| (aire"tior of i \e, ; Ari+ Ari that: 7 from the components: \ - Final Formula Sheet r ' Page 2 Getting components from magrritude and direction: 4 =lllcosl (xcomponent e, =lilsine @ component of 7 ) \ d understood to be the angte that of 7 ) I tr mateswith the positivex axis Projectile Motion e .r direction (motion with constant velocity): a ar=0 t o o v, =vr! X=XO*VxOt y direction (free fall (l) vy = vyo - gt ... positive direction fory taken tobe upward): *|bro *rr)t (2) y = yo (3) y = yo +vrot -)&2 $) "?,=ujo-zsj- n) . Newton's Laws of Motion o Twokindsofforces: contaclforces (objects incontactwithoneanother)andfieldforces (objects not in contact with one another). Gravity is the only field force we will deal with in this course. . t F' = zi (Newton's 2od law) l-/ ximplies two statementr, )4 = ma, atdZr, = *, o Weighf W =mg o Normal forces: in general, whenever two surfaces are in contact, each strface exerts a force on the other in a direction that is perpendicular to the two surfaces. . Friction forces o . Twokinds: theforceof staAcfriction, f, andtheforceofkineticfriction, fr: . -f,3lt,N (force of static friction) . .f* = ltrN (force ofkinetic friction) l, = " the coeffi cient of static fiiction" i p * = " the coeffi cient of kinetic friction" Strings and Springs o o Strings the tension in a sfrng olways pulls m the o$ects attached to the ends of the shing Springs o o o Hooke's fa*' lFl = rt = "the spring constant" or "the force constanf'. Work required to sfietch or compress a spring W o *r. a distancex from its equilibrium length: =!62 2 Potential $ergystored in a spring; U"prns =l*' Formula Sheet.- Final Page 3 Translational Equilibrium . IF=6 + d=d. o The statement immediately abovg really implies two sets of statements: !A . =oand)^$ =o = a,=oa11da, - e. Circular Motion o a* =-,2 'r Centipaalacceleration: Work and Kinetic Energy e o r . Work: W =Fd e,os? Kinetic energy: X -!*u, 2 Work-energy theorem: Wno = [,y Work required to stretch or conpress a spring a distance x from its equilibrium leirgth: w=l k' 2 Potential Energy and Conservative Forces o o Work done by a conservative force: W" - -LU , for some potential energy U Gravitational potential en€rgy (near surface of Earth): (J = mg o o Potential €n€rgy stored in a spring: n* Total mechanical €xrergy: E = U**, =)U' K+U Linear Momentum o o Linearmomeirtum: F= Newton's second ni 9p- ^r (Yr=nm€' \4J u_>O N _./ Implies the following: (I \La r) tqt, - a'P' N (arteragenet force equals averagerate of change of linear momentum). 7=E F)*u a lmpulse: a Impulse-momentum theorem: I = AF Formula Sheet . - Final Collisions o o Two broadcategories: heqd-onand glancing For each category three classes: l. 2. 3. o elastic: I inelastic: andr(conserved f conserved,r(not completely inelastic: f conserved, Knot, objects stick together Head-on collisions: l. elastic: ' frtvti + m2vzi = mtvrf + m2v2l (p-conservatior) t rti * vry =v2; + v2y (bthed' equation..' derived in class) 2. inelastic: €vrenl- t . ntvti+m2v2i=mltf +m2vzf (p*rrrouutLi- )' ;A"/l 3. completelyinelastic: . ntvti * m2v2i = (ry + mzb t (pconservation) o Glancing collisions: ' t . Page 4 Pxi=P4 Pi=Pvl RotationalKinematics o o angle d inradians; 6=! 6=lengthofarcsweptout) angulardispiacernent . ayerageangular r L0=0t -0i velocity: r* =X =# o instantaneousangularvelocity . avsrage angular accelerat io,-r: o instantaneotrs angular acceleration: * o o d understood at= tim{ At+o Ll do, ' Lal - @i Lt -@l tI -ti , = l\# to be in radiansin all of the above Period (timerequired for one complete revolution); ,a f =4 Rotational motion with constant angular acceleration (4 facts): (l) @=@o+dt e) o=on+f,@o+ait (3) 0=eo+aht+!ai (4) at' = ai +2a(0 -0) L + Formula Sheet . - Final Page 5 Rotational and Linear Quantities ' lt = rct) a clr=rd o Centipetal accelerationt a"o - 7412 o igid obiect: Everypoint in the object has the same angular velocity as ev€ry other point in the object. Everypoint in the object has the same angular acceleration as every otler point in the object. . Rolling Without Slipping . v=ra Ypoint , (v:translational ofcontac = 0 speed ofcenter ofrolling object) at each instant Rotational Kinetic Energy and Moment of tnertia ' . General Formula for Moment of Inertia for Collection of.l/Point Masses: =f*,r! i=t e o Rotational Kinetic Energy: X o Total Kinetic Energy: Kaut = Ko*o + K-, Moments of Inertia for Distributed Objects: Table 10-l will be given. *, : !2 tat2 =*.rr'", *f,t"*r' (Note: Here vs* isthehanslationalvelocity ofthecenta ofmass md inqtia about the center of mass.) r t Ig, isthemomentof Torgue and Angular Acceleration o . Torque: t = rF (r is the moment arm) Relation Between Torque and Angular Acceleration (Newton's Second Law for Rotational Motion): to", = Ia . Equilibrium Three conditions required for true equilibrium (translational and rotational equilibrium): l. Zr, =o (i.e., a, =o) 2. Zr, 3. =o trr, =0 (i.e., ay =o) fot any axis (i.e., a=0 about anyais) Formula Sheet ' - Final Page 6 Angular Momentum . o Angular momentum: 7, = Ia . For point particle of mass rz moving with speed v at an angle o 0 with respect to the radial L=rmvsial direction: Newton's Second [.aw for Rotational Motion; tr", Note here that ,ro =* . isthe average net uternal tuqre. t.tiu:.1111.'11.J ..-l.i r f rJi i l-' i il rR: '.t ,... t | .._;t: __- III-r-;-". i l[ i i,i"f /t:{Stir. '{"{;in lft" -.tt ii +ri ti r,l 1i,:,- '/ Disk or solid cvlinder 1i r I r=+riRz (axis at rim) i Ir[il'ff*^ 1 l**u' Longihin rod Disk or solid cylincler I i\L ii \ u-t 'l::i i-l ilr \ "iIL I { jt.a,ri 1. ..r':. - r ...,,,,, Long thin ro Purchase answer to see full attachment