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UW-Madison PHYSICS 207 - Lecture 9

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Page 1Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 1Lecture 9Today:Today: Review sessionAssignment: For Thursday, Read through Chapter 8 (first four sections)Exam Wed., Feb. 17thfrom 7:15-8:45 PM Chapters 1-6One 8½ X 11 hand written note sheet and a calculator (for trig.)Place: Room 2103 (Sec. 301 & 303+) Room 2120 (Sec. 302, TR 2:25-3:15PM, H. Yoo) Physics 207: Lecture 9, Pg 2Textbook Chapters Chapter 1 Concept of Motion Chapter 2 1D Kinematics Chapter 3 Vector and Coordinate Systems Chapter 4 Dynamics I, Two-dimensional motion Chapter 5 Forces and Free Body Diagrams Chapter 6 Force and Newton’s 1stand 2ndLaws (no inclines with friction)Exam will reflect most key points (but not all)25-30% of the exam will be more conceptual70-75% of the exam is problem solvingPage 2Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 3Example The velocity of an object as a function of time is shown in the graph at right. Which graph below best represents the net force vs time relationship for this object?Physics 207: Lecture 9, Pg 4Example The velocity of an object as a function of time is shown in the graph at right. Which graph below best represents the net force vs time relationship for this object?Page 3Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 5Chapter 2Physics 207: Lecture 9, Pg 6Chapter 2Also average speed and average velocityPage 4Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 7Chapter 3Physics 207: Lecture 9, Pg 8Chapter 3Page 5Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 9Chapter 4Physics 207: Lecture 9, Pg 10Chapter 4Page 6Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 11Chapter 5Physics 207: Lecture 9, Pg 12Chapter 5 & 6Page 7Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 13Chapter 6Including static and kinetic friction modelsPhysics 207: Lecture 9, Pg 14Short word problems After breakfast, I weighed myself and the scale read 588 N. On my way out, I decide to take my bathroom scale in the elevator with me. What does the scale read as the elevator accelerates downwards with an acceleration of 1.5 m/s2 ? A bear starts out and walks 1stwith a velocity of 0.60 j m/s for 10 seconds and then walks at 0.40 i m/s for 20 seconds. What was the bear’s average velocity on the walk? What was the bear’s average speed on the walk (with respect to the total distance travelled) ?Page 8Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 15Conceptual ProblemThe pictures below depict cannonballs of identical mass which are launched upwards and forward. The cannonballs are launched at various angles above the horizontal, and with various velocities, but all have the same vertical component of velocity.Physics 207: Lecture 9, Pg 16Conceptual ProblemLet WBand WFbe the weight of the bird and the feeder respectively. Let T be the tension in the wire and N be the normal force of the feeder on the bird. Which of the following free-body diagrams best represents the birdfeeder? (The force vectors are not drawn to scale and are only meant to show the direction, not the magnitude, of each force.) A bird sits in a birdfeeder suspended from a tree by a wire, as shown in the diagram at left.Page 9Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 17Graphing problemThe figure shows a plot of velocity vs. time for an object moving along the x-axis. Which of the following statements is true?(A) The average acceleration over the 11.0 second interval is -0.36 m/s2(B) The instantaneous acceleration at t = 5.0 s is -4.0 m/s2(C) Both A and B are correct.(D) Neither A nor B are correct.Physics 207: Lecture 9, Pg 18Conceptual ProblemA cart on a roller-coaster rolls down the track shown below. As the cart rolls beyond the point shown, what happens to its speed and acceleration in the direction of motion?A. Both decrease.B. The speed decreases, but the acceleration increases.C. Both remain constant.D. The speed increases, but acceleration decreases.E. Both increase.F. OtherPage 10Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 19Conceptual Problem A person initially at point P in the illustration stays there a moment and then moves along the axis to Q and stays there a moment. She then runs quickly to R, stays there a moment, and then strolls slowly back to P. Which of the position vs. time graphs below correctly represents this motion?Physics 207: Lecture 9, Pg 20Sample Problem You have been hired to measure the coefficients of friction for the newly discovered substance jelloium. Today you will measure the coefficient of kinetic friction for jelloium sliding on steel. To do so, you pull a 200 g chunk of jelloium across a horizontal steel table with a constant string tension of 1.00 N. A motion detector records the motion and displays the graph shown.  What is the value of kfor jelloium on steel?Page 11Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 21Sample ProblemΣ Fx=ma = F - ff= F - µkN = F - µkmgΣ Fy= 0 = N – mgµk= (F - ma) / mg & x = ½ a t2 0.80 m = ½ a 4 s2a = 0.40 m/s2 µk= (1.00 - 0.20 · 0.40 ) / (0.20 ·10.) = 0.46Physics 207: Lecture 9, Pg 22Exercise: Newton’s 2ndLawA. 8 x as far B. 4 x as far C. 2 x as far D. 1/4 x as far A force of 2 Newtons acts on a cart that is initially at reston an air track with no air and pushed for 1 second. Because there is friction (no air), the cart stops immediately after I finish pushing. It has traveled a distance, D. Air TrackCartForceNext, the force of 2 Newtons acts again but is applied for 2 seconds. The new distance the cart moves relative to Dis:Page 12Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 23Exercise: SolutionAir TrackCartForce(B) 4 x as longWe know that under constant acceleration, ∆x = a (∆t)2 /2 (when v0=0)Here ∆t2=2∆t1, F2= F1⇒ a2= a1( )4221212121212212=∆∆=∆∆=∆∆tttataxxPhysics 207: Lecture 9, Pg 24Mass-based separation with a centrifugeA centrifuge with a radius of 0.10 m spins at a frequency of 104rpm. How many g’s?Before Afterbb5Page 13Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 25Mass-based separation with a centrifugeHow many g’s?Before Afterar=vt2/ r and vt= ω r = 2π f rvt= (2π x 104 / 60) x 0.10 m/s =100 m/sar= 1 x 104 / 0.10 m/s2 = 10 000 g’sPhysics 207: Lecture 9, Pg 26Another question to ponderHow high will it go?  One day you are sitting somewhat pensively in an airplane seat and notice, looking out the window, one of the jet engines running at full


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UW-Madison PHYSICS 207 - Lecture 9

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