Monday, Oct. 25, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 1 PHYS 1441 – Section 002 Lecture #14 Monday, Oct. 25, 2010 Dr. Jaehoon Yu • Work – Kinetic Energy Theorem Revisited • Work and Energy Involving Kinetic Friction • Potential Energy • Gravitational Potential Energy • Elastic Potential Energy • Mechanical Energy Conservation Today’s homework is homework #8, due 10pm, Tuesday, Nov. 2!!Announcements • Quiz #3 – Class average: 6.6/10 • Equivalent to 66/100 • Previous quizzes: 53/100 and 56/100 – Top score: 10 • 2nd non-comprehensive term exam – Date: Wednesday, Nov. 3 – Time: 1 – 2:20pm in class – Covers: CH3.5 – CH6.7 – There will be a review in class Monday, Nov. 1 • Bring your own problems to solve • Mid-term grade discussion this Wednesday – Dr. Yu’s office, CPB342 • A – F: 12:45 – 1:15 • F – N: 1:15 – 1:50 • N – Y: 1:50 – 2:20 • Colloquium this week – On the subject of renewable energy Monday, Oct. 25, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 2Monday, Oct. 25, 2010 3 PHYS 1441-002, Fall 2010 Dr. Jaehoon YuMonday, Oct. 25, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu Reminder: 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: This Wednesday, Oct. 27 • You must show your OWN, detailed work to obtain any credit!! G = 6.67 × 10−11N ⋅ m2kg2 RE= 6.37 × 103km4Monday, Oct. 25, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 5 When a net external force by the jet engine does work on and object, the kinetic energy of the object changes according to Work-Kinetic Energy Theorem KEf− KEo= 12mvf2−12mvo2Monday, Oct. 25, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 6 The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If a 56.0-mN force acts on the probe parallel through a displacement of 2.42×109m, what is its final speed? Ex. Deep Space 1 = 275 m s( )2+ 2 5.60 × 10-2N( )cos 02.42 × 109m( )474 F∑( )cosθ⎡⎣⎤⎦s =vf= 805 m s 12mvf2−12mvo2Solve for vf vf= vo2+ 2 F∑cosθ( )s mMonday, Oct. 25, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 7 A satellite is moving about the earth in a circular orbit and an elliptical orbit. For these two orbits, determine whether the kinetic energy of the satellite changes during the motion. Ex. Satellite Motion and Work By the Gravity For a circular orbit For an elliptical orbit No change! Why not? Gravitational force is the only external force but it is perpendicular to the displacement. So no work. Changes! Why? Gravitational force is the only external force but its angle with respect to the displacement varies. So it performs work.Monday, Oct. 25, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 8 Work and Energy Involving Kinetic Friction • What do you think the work looks like if there is friction? – Static friction does not matter! Why? – Then which friction matters? M M d vi vf Friction force Ffr works on the object to slow down The work on the object by the friction Ffr is Wfr=The final kinetic energy of an object, taking into account its initial kinetic energy, friction force and other source of work, is Ffr t=0, KEi Friction, Engine work t=T, KEf It isn’t there when the object is moving. Ffrd cos 180( )= − Ffrd − Ffrd − FfrdKinetic Friction The negative sign means that the work is done on the friction!!Monday, Oct. 25, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 9 What are the forces in this motion? Ex. Downhill Skiing A 58kg skier is coasting down a 25o slope. A kinetic frictional force of magnitude fk=70N opposes her motion. At the top of the slope, the skier’s speed is v0=3.6m/s. Ignoring air resistance, determine the speed vf at the point that is displaced 57m downhill. Gravitational force: Fg Normal force: FN Kinetic frictional force: fk What are the X and Y component of the net force in this motion? Y component From this we obtain What is the coefficient of kinetic friction?Monday, Oct. 25, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 10 Ex. Now with the X component X component Total work by this force From work-kinetic energy theorem Solving for vf mg sin 25− fk( )⋅ s =What is her acceleration?Monday, Oct. 25, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 11 Example of Work Under Friction A 6.0kg block initially at rest is pulled to East along a horizontal surface with coefficient of kinetic friction mk=0.15 by a constant horizontal force of 12N. Find the speed of the block after it has moved 3.0m. Work done by the force F is Thus the total work is WF=M F M d=3.0m vi=0 vf Work done by friction Fk is Fk Wk=Fk⋅d == 0.15 × 6.0 × 9.8 × 3.0 cos180 = −26 J( )Using work-kinetic energy theorem and the fact that initial speed is 0, we obtain Solving the equation for vf, we obtain µkmg dcosθ Fd cosθ=Monday, Oct. 25, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 12 Potential Energy Energy associated with a system of objects Stored energy which has the potential or the possibility to work or to convert to kinetic energy What does this mean? In order to describe potential energy, U, a system must be defined. What are other forms of energies in the universe? The concept of potential energy can only be used under the special class of forces called the conservative force which results in the principle of conservation of mechanical energy. Mechanical Energy Biological Energy Electromagnetic Energy Nuclear Energy Chemical Energy These different types of energies are stored in the universe in many different forms!!! If one takes into account ALL forms of energy, the total energy in the entire universe is conserved. It just transforms from one form to another. Thermal EnergyMonday, Oct. 25, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 13 Gravitational Potential Energy When an object is falling, the gravitational force, Mg, performs the work on the object, increasing the object’s kinetic energy. So the potential energy of an object at a height y,the potential to do work, is expressed as This potential energy is given to an object by the gravitational field in the system of Earth by virtue of the object’s height from an
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