X2.0 205PHYS-205(10) (Kaldon-18454) Name __________________________________________WMU-Winter 2002 Exam 0 - 100,000 points + 20,000 ¶ points Book Title ________This is for Topic 1, not your textbook!___Sample – Not a Real Exam Rev. 09/29/2000.2State Any Assumptions You Need To Make – Show All Work – Circle Any Final AnswersUse Your Time Wisely – Work on What You Can – Be Sure to Write Down EquationsFeel Free to Ask Any Questions¶2a ¶2b ¶2c ¶2eREVENGE OF THE GODS – PART I (25,000 points)1.) In Greek mythology, Sisyphus was condemned by Zeus to roll a bigstone up a hill, only to have it roll down the hill every time. What youwould never have learned in school is that this was Zeus’ secondattempt at punishment. The first time, he made Sisyphus push a greatblock of stone ( m = 250. kg ) up a ramp 10.4 m high with a 30° incline. The coefficients of friction are mk= 0.700 and ms = 0.900. (a) Draw the free body diagram of the block of stone, labeling everything ofinterest correctly.(b) Find the normal force acting on the stone block.(c) Find the force of friction as Sisyphus is pushing the stone block up the inclined ramp at 0.875 m/s.(d) Find the magnitude and direction of the force that Sisyphus pushes on the stone block.(e) Show that when Sisyphus gets the block to the top of the ramp and stops, that the stone block does notslide back down. “Oops,” says Zeus. “Back to the drawing board.”Physics 205 / Sample Exam 2 Winter 2002 Page 2Odds and Ends (25,000 points)2.) An object of mass m = 4.00 kg begins its motion at x0 = 4.00 m, v0 = 4.00 m/s, a0 = 4.00 m/s² andan initial jerk of j0 = 4.00 m/s³. The motion of an object is determined by the followingequation: ¶(a) Find the equation for the force, F, acting on this object.¶(b) In the strange world of Dr. Seuss, a fandoogle is a device that had a force, FxxxNmNm=+400 40044..Find the work done by this force from x = 0 to x = 4.00 m along y = 0.¶(c) Find the work done by the fandoogle force from y = 0 to y = 4.00 m along x = 4.00 m.(d) Find the work done when rFN= 500.$$i + 6.00N j and the displacement is 3.85 m @ 30°.¶(e) An object of mass 71.3 kg has a motion that follows the following equations. Find the vector forcerF at time t = 0.xtmmsmsmsmsms()../././././=+++++4 00 4 00 4 00 4 00 4 00 4 0022 33 77 88tt t t tyt m ms ms() . . / . /=+ +400 400 40022ttdxdtms444400= ./Physics 205 / Sample Exam 2 Winter 2002 Page 3This Part of the Test is a Drag (25,000 points)3.) A truck weighing 66,000 pounds (m = 30,000 kg) tries to get moving on a road that is covered in sheetice. The coefficients of friction of rubber tires on ice are 0.15 and 0.20 respectively. (a) What is themaximum acceleration that the truck can have, assuming the tires make good contact with the road?(b) At a speed of 8.92 m/s, the truck driver feels the truck beginning to fishtail and stomps on the brakes,causing all 18 truck tires to lose good contact with the road. Find the distance it takes for the truck tocome to a stop.(c) Identify the Newton’s Third Law force connected to the friction force in (b). Short answer.(d) The air resistance force on a particular falling object (m = 1.753 kg) can be written as FBv=2, whereB = 12.0 N·s²/m². Find the terminal velocity of the object. Hint: As usual, start with the Free BodyDiagram…(e) The last force we had that included a term v², was the centripetal force. Briefly explain why, v doesn’tchange, if there is a net centripetal force acting on a body that is undergoing Uniform Circular Motion,and since F = ma by Newton’s Second Law.Physics 205 / Sample Exam 2 Winter 2002 Page 4When It Absolutely Positively Has To Fall (25,000 points)4.) Jill from FedEx™ is riding the elevator to deliver some packages. The loaded elevator has a mass of505 kg. (a) If the elevator is just sitting there, find the tension in the cable that pulls up on the elevator.(b) Find the work done in raising the elevator 20.0 m, using the tension in thecable as the force. If you didn’t get an answer to (a), use T = 505. N.(c) The elevator is raised 20.0 m. What is its change in potential energy, DU?(d) The elevator needs to go up 20.0 m in 10.0 seconds. Neglecting any accelerations to start or stop theelevator, how much power is needed to do this work?(e) If the elevator is accelerating down at 2.23 m/s², find the tension in the cable.Physics 205 / Sample Exam 2 Winter 2002 Page 5Pyramid Power (25,000 points)5.) It is speculated that the Great Pyramids of Egypt were built by sliding blocks of stone along massivesloping ramps. Let us examine the Physics of moving a 5000. kg blockof stone from the ground up to the top of The Great Pyramid ofCheops, the largest ever built at 482 ft (147 m) high. The coefficientsof friction are mk = 0.700 and ms = 0.900. (a) Find the force required topush the block up a 30° incline.(b) Find the force required to push the block up a 60° incline.(c) Find the work required to push the block up a 30° incline.(d) Find the work required to push the block up a 60° incline.(e) If the block moves at a constant speed of 0.100 m/s, find the power needed to push the block up the30° incline and the 60° incline. Would you expect these two powers to be the same?Physics 205 / Sample Exam 2 Winter 2002 Page 6Odds and Ends (25,000 points)6.) ¶(a) A car (m = 2300. kg) moving at v0 = 35.0 m/s crashes into the side of a mountain. The force tocrush the car is F = C x³, where x is the how far the car is crushed. If the car ends up 0.500 m shorter, findthe value of the constant C. Hint: Don’t know about you, but I’d use Work and Energy here.¶(b) U = mgy is the expression for the work done by gravity near the surface of the Earth. IntegrateFGm mr=122 to find the work done by gravity anywhere.¶(c) Find the work done when rFN= 500.$$i + 6.00N j and the displacement is 3.85 m @ 30°.(d) Car No 1 ( m = 2000. kg ) is moving to the right at v = 25.0 m/s. Car No 2 ( m = 3000. kg ) is movingto the left at v = 15.0 m/s. If they collide totally inelastically, find vx of the wreck.¶(e) An object of mass 2.25 kg has a motion that follows the following equations. Find the vector forcerF at time t = 0.xtmmsmsmsmsms()../././././=+ + + + +500 500 500 500 500 50022 33 44 55tt t t tyt m ms ms() . . / . /=+ +500 500 50022ttPhysics 205 / Sample Exam 2 Winter 2002 Page 7 “Ground Control to Major Tom…” (25,000 points)7.) Last week