Phys 201 1nd EditionExam # 2 Study Guide Lectures: 8 - 12Lecture 8 (September 29)A 25.00 kg block is pushed up a frictionless inclined plane that makes an angle of θ = 27 o with the horizontal direction. What force is needed to move the block with constant speed? Assume that the force is parallel to the surface of the plane.Answer: 111.3NExplanation: First step is to make a free body diagram.F N A B mgBecause weight goes straight down, we can make a hypothetical triangle using the weight (mg) as the y component. This means that side b of the hypothetical triangle will be equal to the force moving the block.Weight=mg =(25)(9.81)= 245.25NF=B=Cos(27)mg =245.25NSin(27)=111.34NLecture 8 (September 29)What is the minimum force required to push a 40 kg bag with a coefficient of friction of 0.32 across the floor?Answer: 125.6N27◦Explanation: All we have to do for this problem is plug in our know values to solve for force. Remember that the F=μN where μ represents the friction coefficient and N is equal to the weight of an object.F=μN = μmgF=(.32)(40)(9.81)= 125.57NLecture 8 (September 29)A 0.840- kg glider on a level air track is joined by strings to two hanging masses. The mass on the left is 4.85 kg and the one on the right is 3.62 kg The strings have negligible mass and pass over light, frictionless pulleys. A) Find the acceleration of the masses when the air flow is turned off and the coefficient of friction between the glider and the track is 0.43. B) Find the tension between the glider and the mass on the right. C) Find the tension between the glider and the mass on the left. Take positive direction to be acceleration to the right. Answer: A)-.915m/s2 B) 38.8N C) 43.1NExplanation: Make three separate free body diagrams for each of the three masses;Right mass: T1mgMiddle mass: T2 T1 +FfT2Let mass: mgTo find the acceleration, you need to make equations for each of the three objects separately and then combine them. You need to remember that forces pulling to the left are negative, and you need to remember that the force of friction (Ff) is the friction coefficient(.43) times the normal force, which would be equal to the weight. If you do 3 separate free body diagrams where T1 and T2 represent the two different tension forces, you should come up with these equations;right mass: Fnet= 3.62g - T1= ma = 3.62aMiddle mass: Fnet=T1-(T2-Ff)=Ff+T1-T2=ma=.84aleft mass: Fnet=T2-4.85g=ma =4.85aOnce you have these equations, isolate T1 and T2 for the right and left masses.Right mass: T1=3.62g-3.62aLeft mass: T2=4.85g + 4.85aNow plug these values into the equation for the middle mass, then simplify the equation so thatyou're solving for aFf+(3.62g-3.62a)-(4.85g+4.85a)=.84aFf+3.62g-4.85g=.84a + 4.85a + 3.62aFf + 3.62g -4.85g=9.31aNow just plug in your values and divide to solve for a(.84*9.81*.43)-(4.85*9.81)+(3.62*9.81)=9.31aKeep the acceleration negative. Since the mass to the left is heavier it makes sense that our acceleration is in the negative direction.To find the tension for the right and left masses just plug this value for acceleration into the equations you got for the right and left masses.Lecture 9 (October 2)A velodrome track is banked so that a bicycle traveling at 63.6 km/h will have no tendency to slip to either side when traveling on the path that has a radius of curvature of 60.0 m. What is the banking angle?Answer: 27.94◦Explanation: For any problem where you have velocity, radius, and banking angle, but no force of friction, use the equation Tan(θ)=(V2/r)/g. Essentially you're looking at this in terms of acceleration, angular velocity, and net force. Because there's no friction, we're looking at the force in the x direction which is making the car go, and the force in the y direction, which is essentially the weight. So the equations for these two forces would be Fx=ma=m*V2/rFy=mgIf you do a free body diagram of the circle, the vector Fx would be on the outside of the circle, which makes sense because the movement is in the x direction, and the movement is taking place around the circle. The Fy vector would connect the center of the circle to the bottom of the Fx vector. This means that if we made a triangle out of these vectors, Fx would beopposite of the banking angle, and Fy would be adjacent to it. This means to solve for the angle we would do the Tan(θ)=opp/adjTan(θ)=(m*V2/r)/(mg)--> the m's cancel out so you're left with Tan(θ)=(V2/r)/g, which is just.Before you plug in your values remember to convert the velocity from km/hr to m/s. 63.6km/hr= 63600m/3600s= 17.67m/s Tan(θ)=(V2/r)/g --> Tan(θ)=(17.672/60)/9.81=.5303Tan-1(.5303)Lecture 9 (October 2)A spring scale on a rotating platform indicates that the horizontal force on a 0.496 kg mass is 1.32 N when the mass is 2.08 m from the axis of rotation. How long does it take for the platformto make one revolution?Answer: 5.56sExplanation: You're trying to find the period. First thing you need to do is find the acceleration from the force divided by the mass. Then you can use acceleration and the radius that they give you to find velocity, which you can use that to find the period. If you remember that whole mess of equations from the lecture on circular motion that he said were all equivalent to on another, this is where they come in handy.F=ma --> a=1.32/.496= 2.66a=V2/r --> 2.66m/s2=V2/2.08 --> V= √(2.66*2.08) = 2.35m/sV/r = 2π/T --> T= 2π/(V/r) ---> T= 2π/(2.35/2.08)= 5.56sLecture 9 (October 2)A coin placed on a turntable rotating at 31.4 rev/min will stay there if its center is placed no further than 8.9 cm from the axis of rotation. A) What is the maximum distance the center of the coin can be placed from the axis if the turntable rotates at 45.1 rev/min? B) What is the coefficient of friction between the coin and turntable?Answer: A) 4.3cm B) .982Explanation: a) You can figure out V by using the equations V/r=2πf. f is the frequency, which is your rev/minute. Make sure you divide the revolutions by 60 and for r convert to meters, since V has to be in m/s. V/r=2πf --> V=2πf*r --> V=2π(31.4/60)*(.089m)= .2926Now plug V into the equation V2/r=g*Tan(θ). V2/r=g*Tan(θ) --> (V2/r)/g=Tan(θ) --> Tan(θ)=(.29262/.089)/9.81 = .098When you have Tan(θ) you can find r by modifying the equation to be r=(2πf*r)2/(g*Tan(θ)). Remember to convert