Physics Exam 205 - Winter 2002 Sample Exam - Prof. Philip E. Kaldon, Exams of Physics

Download Physics Exam 205 - Winter 2002 Sample Exam - Prof. Philip E. Kaldon and more Exams Physics in PDF only on Docsity! X2.0 205 PHYS-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. 09/29/2000.2 State Any Assumptions You Need To Make – Show All Work – Circle Any Final Answers Use Your Time Wisely – Work on What You Can – Be Sure to Write Down Equations Feel Free to Ask Any Questions
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2e REVENGE OF THE GODS – PART I (25,000 points) 1.) In Greek mythology, Sisyphus was condemned by Zeus to roll a big stone up a hill, only to have it roll down the hill every time. What you would never have learned in school is that this was Zeus’ second attempt at punishment. The first time, he made Sisyphus push a great block of stone ( m = 250. kg ) up a ramp 10.4 m high with a 30° incline. The coefficients of friction are
k = 0.700 and
s = 0.900. (a) Draw the free body diagram of the block of stone, labeling everything of interest 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 not slide back down. “Oops,” says Zeus. “Back to the drawing board.” Physics 205 / Sample Exam 2 Winter 2002 Page 2 Odds 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 and an initial jerk of j0 = 4.00 m/s³. The motion of an object is determined by the following equation:
(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, F x xx N m N m
4 00 4 004 4. . 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
F N
5 00.  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 force
F at time t = 0. x t m m s m s m s m s m s( ) . . / . / . / . / . /
    4 00 4 00 4 00 4 00 4 00 4 002 2 3 3 7 7 8 8t t t t t y t m m s m s( ) . . / . /
 4 00 4 00 4 00 2 2t t d x dt m s 4 4 44 00
. / Physics 205 / Sample Exam 2 Winter 2002 Page 3 This 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 sheet ice. The coefficients of friction of rubber tires on ice are 0.15 and 0.20 respectively. (a) What is the maximum 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 to come 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 F Bv
2 , where B = 12.0 N·s²/m². Find the terminal velocity of the object. Hint: As usual, start with the Free Body Diagram… (e) The last force we had that included a term v², was the centripetal force. Briefly explain why, v doesn’t change, 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 4 When 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 of 505 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 the cable 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, U? (d) The elevator needs to go up 20.0 m in 10.0 seconds. Neglecting any accelerations to start or stop the elevator, 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.