Physics 211 Exam #2 Study Guide: Lectures 8-18Lecture 8 (September 15th)I. Projectile Motion***A projectile launched in an arbitrary direction may have initial velocity components in both x and y directions but it is still accelerated downward***Note that velocity in the x direction does not change because the components of velocity in the x and y direction are independent of each otherII. Example Problems for Projectile MotionD) they all have the same hang time because all of them reach the same heightPhysics 211 Exam #2 Study Guide: Lectures 8-18E) The cannonball on the moon will be accelerated downward at a rate of approximately 1/6 of the rate of gravity on earthLecture 9 (September 19th)I. What is Force? ΣF = ma-air pressure, mass times acceleration, acceleration due to gravity, tension, push, pull, tug-a force is something that is capable of changing an object’s state of motion (its velocity)II. Inertia- the natural tendency of an object to remain at rest or in constant velocity-ExampleWeight = how much it pushes down on your handInertia = how difficult it is to move from side to sideLecture 10 (September 22nd)I. Newton’s 2nd Law*Force equals mass times acceleration Weight = mass x gravity = mg Fnet = ma Force is expresses in Newtons, a Newton = (kg)(m/s2)Physics 211 Exam #2 Study Guide: Lectures 8-18If a car pushes an object with a mass of 390 kg that is moving at 6m/s in 4 seconds, what is the net force exerted on the object?a = (6m/s)/4s = 1.5 m/s2 Fnet = ma Fnet = (390kg)(1.5m/s2) = 292NLecture 11 (September 24th)I. Newton’s 2nd Law Practice ProblemsA force acts on a 1.5 kg mass, giving it an acceleration of 3m/s2. If the same force acts on a 2.25 kg mass, what accleration will be produced?F1=m1a1 F2=m2a2 = (3kg)(1.5m/s2) a = F2/m2 = 4.5N = (4.5N)/(2.25kg) = 2.0m/s2Two masses are attached to a string and on a pulley. Write the equations for the net force in terms of the components. y = ∑F = Fnet FN-m1g(cosθ) = m1ay N = m1g(cosθ)x = ∑F = Fnet T = m1g(sinθ) = m1ax Break down the force into x and y components x = ∑F = Fnet T2sinθ – T = ma T1 = T2sinθPhysics 211 Exam #2 Study Guide: Lectures 8-18y = ∑F = Fnet T2cosθ – T1 = ma T2 = T1/cosθLecture 12 (September 25th)I. Practice ProblemA father pushes his 25 kg daughter with a force of 200 N at an angle of 30 dgrees below the horizontal. What is the acceleration? What is the normal force?Fcos = x = 200cos30 = 173NƟFsin = y = 200sin30 = 100NƟMg = (25kg)(9.8M/S2)= 245 nFpx = maxAx = fpx/m = 173N/25kg = 6.9m/s2Y:ΣF = FnetN-mg-fpy = mayN = mg +Fpy = 245N +100N = 345 NII. Newton’s 3rd LawFor every action there is an equal and opposite reactionFA on B = - FB on ALecture 13 (September 26th)iClicker Review: No notesLecture 14 (September 29th)I. Friction Always opposes direction of motionTypes of frictionStatic (μs): when the frictional force is large enough to prevent motionKinetic (μk): when two surfaces are sliding along each otherCase 1Case 2Physics 211 Exam #2 Study Guide: Lectures 8-18Rolling friction: When an object is rolling without slippingII. Friction Practice ProblemsA 70 kg filing cabinet needs to be slid across a floor if the static coefficient of friction is 0.5 and the kinetic coefficient is 0.4, how hard to you have to push to move it at a constant velocity?Y: ∑F = FnetN-mg = maN = mg = (70kg)(9.8m/s2) X: ∑F = FnetFp- F = maFp = fFk = μkN = (.4)(686N) = 274 NFs = μsN = (.5)(680N) = 343 NYou need to push with 343 N to get it started and then contuinue to push at 274 N to keep it moving.If you are sledding on a hill with a slope of 30 degrees with a constant velocity, what is the coefficient of friction?Y: ∑F = FnetN-mgcosθ = maN = mgcosθX: ∑F = FnetMgsinθ –fk = maMgsinθ = fkMk = sinθ/cosθ = OH/AH = tanθMk = tan30 = .507Lecture 15 (October 1st)I. WorkWhat is work?Work done by a constant force:Physics 211 Exam #2 Study Guide: Lectures 8-18Lecture 16 (October 2nd)I. Work done by a variable forceExample: the force exerted by a spring varies linearly with the displacementk = spring constant, x = distanceII. The Work-Energy TheoremThis relationship is called the work-energy theoremPhysics 211 Exam #2 Study Guide: Lectures 8-18Equations we can use as a result of the work-energy theorem: A .25 kg puck is acted upon by a stick with a force of 6N for .5m. What is its kinetic energy? How much work is needed to stop it?GPEi + KEi + W = GPEf + KefW = Fd = (6N)(.5m) = 3JLecture 17 (October 3rd)I. Work Practice ProblemsA 30g bullet moving 350m/s hits a tree and stops in 12cm. What force is needed to stop the bullet?GPEi + KEi + W = GPEf + KEfa = (-350 m/s) 2 = 510,000m/s2 2(.012m)W = -KE = 1/2 mv2 = -184 JF = w/d = 1531 NA soda has 170 calories. How far would I need to walk to have the same potential energy?170 Calories = (10,000calories/1 Calorie) = 711,000 JGPEi + KEi + W = GPEf + KEfW = GPE = mgh h = 711,000J/(75kg)(9.8m/s2) = 968mLecture 18 (October 6th)I. Work ProblemsA 70g pinball spring is compressed 8cm. What’s the final velocity of the ball?W = FdGPEi + KEi + W = GPEf + KEfPhysics 211 Exam #2 Study Guide: Lectures 8-18EPEi = ½ kx2 = 9.6JGPEf = .01JVf2 = 2EPEi/m V = 16.5m/sA .05m pendulum is pulled 35 degrees to the side. How fast is it going at the bottom? Mghi = ½ mvf2Vf = square root of 2ghi = 1.33m/sII. PowerAverage power is the total amount of work done divided by the time takenP = W/tWhat power is needed to move a 2,000kg elevator at 3m/s?P = W/t W=mgh=(2000kg)(4.8m/s2)(3m)P =