PHYS 1441 – Section 002 Lecture #13AnnouncementsSlide 3Reminder: Special ProjectMotion in Resistive ForcesWork Done by a Constant ForceLet’s think about the meaning of work!Work done by a constant forceScalar Product of Two VectorsExample of Work by Scalar ProductEx. Pulling A Suitcase-on-WheelEx. 6.1 Work done on a crateEx. Bench Pressing and The Concept of Negative WorkEx. Accelerating a CrateEx. Continued…Kinetic Energy and Work-Kinetic Energy TheoremWednesday, Oct. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu1PHYS 1441 – Section 002Lecture #13Wednesday, Oct. 20, 2010Dr. Jaehoon Yu•Motion in Resistive Force•Work done by a constant force•Scalar Product of the Vector•Work with friction•Work-Kinetic Energy Theorem•Potential EnergyAnnouncements•2nd non-comprehensive term exam–Date: Wednesday, Nov. 3–Time: 1 – 2:20pm in class–Covers: CH3.5 – what we finish Monday, Nov. 1•Physics faculty research expo todayWednesday, Oct. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu2Wednesday, Oct. 20, 2010 3PHYS 1441-002, Fall 2010 Dr. Jaehoon YuWednesday, Oct. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon YuReminder: Special Project•Using the fact that g=9.80m/s2 on the Earth’s surface, find the average density of the Earth.–Use the following information only•The gravitational constant •The radius of the Earth•20 point extra credit•Due: Wednesday, Oct. 27•You must show your OWN, detailed work to obtain any credit!! G =6.67×10−11N ⋅m2kg2 RE=6.37×103km4Wednesday, Oct. 20, 2010Motion in Resistive ForcesMedium can exert resistive forces on an object moving through it due to viscosity or other types frictional properties of the medium.These forces are exerted on moving objects in opposite direction of the movement. Some examples? These forces are proportional to such factors as speed. They almost always increase with increasing speed. Two different cases of proportionality: 1. Forces linearly proportional to speed: Slowly moving or very small objects2. Forces proportional to square of speed: Large objects w/ reasonable speedAir resistance, viscous force of liquid, etcPHYS 1441-002, Fall 2010 Dr. Jaehoon Yu5Wednesday, Oct. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu6xyWork Done by a Constant ForceA meaningful work in physics is done only when the net forces exerted on an object changes the energy of the object.MFθFree Body DiagramMdgF M g=ur urNFurθFurWhich force did the work?W =Force FurHow much work did it do?What does this mean?Physically meaningful work is done only by the component of the force along the movement of the object.Unit?N m�Work is an energy transfer!! Fur∑( )⋅dur=cosFd q (for Joule)J=Why? What kind? ScalarLet’s think about the meaning of work!•A person is holding a grocery bag and walking at a constant velocity.•Is he doing any work ON the bag? –No–Why not?–Because the force he exerts on the bag, Fp, is perpendicular to the displacement!!–This means that he is not adding any energy to the bag.•So what does this mean?–In order for a force to perform any meaningful work, the energy of the object the force exerts on must change!! •What happened to the person?–He spends his energy just to keep the bag up but did not perform any work on the bag.Wednesday, Oct. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu7Wednesday, Oct. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu8W =cos 0 =oWork done by a constant force = F cosθ( )scos90 =ocos180 =o101-s rF ⋅rsWednesday, Oct. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu9Scalar Product of Two Vectors •Product of magnitude of the two vectors and the cosine of the angle between them Aur⋅Bur≡•Operation is commutative Aur⋅Bur= AurBurcosθ =•Operation follows the distribution law of multiplication Aur⋅Bur•How does scalar product look in terms of components? Aur=Axi∧+Ayj∧+Azk∧ Bur=Bxi∧+Byj∧+Bzk∧ Aur⋅Bur= Axi∧+Ayj∧+Azk∧⎛⎝⎜⎞⎠⎟⋅ Bxi∧+Byj∧+Bzk∧⎛⎝⎜⎞⎠⎟x x y y z zA B i i A B j j A B k k� � � � � �� �= �+ �+ �� �� �kkjjii ikkjji•Scalar products of Unit Vectors Aur⋅Bur= AurBurcosθ BurAurcosθ = Bur⋅ Aur Aur⋅ Bur+Cur( )=10x xA B +y yA B +z zA B=0 cross terms+ +Aur⋅CurWednesday, Oct. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu10Example of Work by Scalar ProductA particle moving on the xy plane undergoes a displacement d=(2.0i+3.0j)m as a constant force F=(5.0i+2.0j) N acts on the particle. a) Calculate the magnitude of the displacement and that of the force. dur= Fur=b) Calculate the work done by the force F.Wjiji 0.20.50.30.2)(166100.20.30.50.2 Jjjii YXdFCan you do this using the magnitudes and the angle between d and F?W =22yxdd m6.30.30.22222yxFF N4.50.20.522 Fur⋅dur= Fur⋅dur= FurdurcosθWednesday, Oct. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu11Ex. Pulling A Suitcase-on-WheelFind the work done by a 45.0N force in pulling the suitcase in the figure at an angle 50.0o for a distance s=75.0m.Does work depend on mass of the object being worked on?W = = 45.0⋅cos50o( )⋅75.0=2170JYesWhy don’t I see the mass term in the work at all then?It is reflected in the force. If an object has smaller mass, it would take less force to move it at the same acceleration than a heavier object. So it would take less work. Which makes perfect sense, doesn’t it? Fur∑( )⋅dur= Fur∑( )cosθ durWednesday, Oct. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu12Ex. 6.1 Work done on a crateA person pulls a 50kg crate 40m along a horizontal floor by a constant force Fp=100N, which acts at a 37o angle as shown in the figure. The floor is rough and exerts a friction force Ffr=50N. Determine (a) the work done by each force and (b) the net work done on the crate. What are the forces exerting on the crate? WG=FG=-mgSo the net work on the crate Wnet=Work done on the crate by FGFpFfrWhich force performs the work on the crate?FpFfr Wp=Work done on the crate by Fp:Work done on the crate by Ffr: Wfr=This is the same as Wnet= rFG⋅rx = −mgcos −90o( )⋅rx = 0J Furp⋅rx = Furpcos37o⋅rx = 100 ⋅cos37o⋅40=3200J Furfr⋅rx = Furfrcos180o⋅rx = 50 ⋅cos180o⋅40=−2000J
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