1. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 1 PHYS 1441 – Section 001 Lecture #12 Tuesday, June 24, 2014 Dr. Jaehoon Yu Work done by a constant force Multiplication of Vectors Work-Kinetic Energy Theorem Work and Energy Involving Kinetic Friction Potential Energy Today’s homework is homework #7, due 11pm, Friday, June 27!!

2. Announcements Term exam #2 –In class this Wednesday, June 25 –Non-comprehensive exam –Covers CH 4.7 to what we finish today, Tuesday, June 24 –Bring your calculator but DO NOT input formula into it! Your phones or portable computers are NOT allowed as a replacement! –You can prepare a one 8.5x11.5 sheet (front and back) of handwritten formulae and values of constants for the exam  no solutions, derivations or definitions! No additional formulae or values of constants will be provided! Quiz 3 results –Class average: 27.4/45 Equivalent to: 61/100 Previous results: 69.3/100 and 59.6/100 –Top score: 45/45 Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 2

3. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 3 x y Work Done by a Constant Force A meaningful work in physics is done only when the net forces exerted on an object changes the energy of the object. M F  Free Body Diagram M d  Which force did the work?Force How 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? Work is an energy transfer!! Why? What kind?Scalar

4. Let’s think about the meaning of work! A person is holding a grocery bag and walking at a constant velocity. Are his hands doing any work ON the bag? –No –Why not? –Because the force hands exert on the bag, Fp,Fp, is perpendicular to the displacement!! –This means that hands are not adding any energy to the bag. So what does this mean? –In order for a force to have performed any meaningful work, the energy of the object the force exerts on must change due to that force!! What happened to the person? –He spends his energy just to keep the bag up but did not perform any work on the bag. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 4

5. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 5 Work done by a constant force s

6. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 6 Scalar Product of Two Vectors Product of magnitude of the two vectors and the cosine of the angle between them Operation is commutative Operation follows the distribution law of multiplication How does scalar product look in terms of components? Scalar products of Unit Vectors =0

7. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 7 Example of Work by Scalar Product A particle moving on the xy plane undergoes a displacement d =(2.0 i +3.0 j )m as a constant force F =(5.0 i +2.0 j ) N acts on the particle. a) Calculate the magnitude of the displacement and that of the force. b) Calculate the work done by the force F. Y X d F Can you do this using the magnitudes and the angle between d and F?F?

8. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 8 Ex. Pulling A Suitcase-on-Wheel Find the work done by a 45.0N force in pulling the suitcase in the figure at an angle 50.0 o for a distance s=75.0m. Does work depend on mass of the object being worked on?Yes Why 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?

9. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 9 Ex. 6.1 Work done on a crate A person pulls a 50kg crate 40m along a horizontal floor by a constant force F p =100N, which acts at a 37 o angle as shown in the figure. The floor is rough and exerts a friction force F fr =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? F G =-mg So the net work on the crate Work done on the crate by FGFG FpFp F fr Which force performs the work on the crate? FpFp F fr Work done on the crate by Fp:Fp: Work done on the crate by F fr : This is the same as F N =+mg Work done on the crate by FNFN

10. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 10 Ex. Bench Pressing and The Concept of Negative Work A weight lifter is bench-pressing a barbell whose weight is 710N a distance of 0.65m above his chest. Then he lowers it the same distance. The weight is raised and lowered at a constant velocity. Determine the work in the two cases. What is the angle between the force and the displacement? What does the negative work mean? The gravitational force does the work on the weight lifter!

11. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 11 Kinetic Energy and Work-Kinetic Energy Theorem Some problems are hard to solve using Newton’s second law –If forces exerting on an object during the motion are complicated –Relate the work done on the object by the net force to the change of the speed of the object M ΣF M s vivi vfvf Suppose net force ΣF ΣF was exerted on an object for displacement d to increase its speed from vi vi to vf.vf. The work on the object by the net force ΣF ΣF is Using the kinematic equation of motion Work Kinetic Energy Work done by the net force causes change in the object’s kinetic energy. Work-Kinetic Energy Theorem

12. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 12 When a net external force by the jet engine does work on an object, the kinetic energy of the object changes according to Work-Kinetic Energy Theorem

13. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 13 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 2.42×10 9 m, what is its final speed? Ex. Deep Space 1 Solve for v f

14. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 14 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.

15. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 15 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 vivi vfvf Friction force F fr works on the object to slow down The work on the object by the friction F fr is The final kinetic energy of an object, including its initial kinetic energy, work by the friction force and all other sources of work, is F fr t=0, KE i Friction, Engine work t=T, KE f It isn’t there when the object is moving. Kinetic Friction The negative sign means that the work is done on the friction!!

16. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 16 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  k =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 M F M d=3.0m v i =0 vfvf Work done by friction F k is FkFk Using work-kinetic energy theorem and the fact that initial speed is 0, we obtain Solving the equation for v f, we obtain

17. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 17 What are the forces in this motion? Ex. Downhill Skiing A 58kg skier is coasting down a 25 o slope. A kinetic frictional force of magnitude f k =70N opposes her motion. At the top of the slope, the skier’s speed is v 0 =3.6m/s. Ignoring air resistance, determine the speed vf vf at the point that is displaced 57m downhill. Gravitational force: FgFg Normal force: FNFN Kinetic frictional force: fkfk x y 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?

18. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 18 Ex. Now with the X component X component Total work by this force From work-kinetic energy theorem Solving for vfvf What is her acceleration?

19. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 19 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 EnergyBiological Energy Electromagnetic EnergyNuclear 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 Energy

20. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 20 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 arbitrary zero level m hfhf m mgmg hihi What does this mean? The work done on the object by the gravitational force as the brick drops from y i to y f is: Work by the gravitational force as the brick drops from y i to yfyf is the negative change of the system’s potential energy  Potential energy was spent in order for the gravitational force to increase the brick’s kinetic energy. (since )

21. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 21 A gymnast leaves the trampoline at an initial height of 1.20 m and reaches a maximum height of 4.80 m before falling back down. What was the initial speed of the gymnast? Ex. A Gymnast on a Trampoline

22. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 22 Ex. Continued From the work-kinetic energy theorem Work done by the gravitational force Since at the maximum height, the final speed is 0. Using work-KE theorem, we obtain

23. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 23 Example for Potential Energy A bowler drops bowling ball of mass 7kg on his toe. Choosing the floor level as y=0, estimate the total work done on the ball by the gravitational force as the ball falls on the toe. b) Perform the same calculation using the top of the bowler’s head as the origin. Assuming the bowler’s height is 1.8m, the ball’s original position is –1.3m, and the toe is at –1.77m. M Let’s assume the top of the toe is 0.03m from the floor and the hand was 0.5m above the floor. What has to change? First we must re-compute the positions of the ball in his hand and on his toe.

24. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 24 Elastic Potential Energy The force spring exerts on an object when it is distorted from its equilibrium by a distance x is Potential energy given to an object by a spring or an object with elasticity in the system that consists of an object and the spring. What do you see from the above equations? The work performed on the object by the spring is The work done on the object by the spring depends only on the initial and final position of the distorted spring. Where else did you see this trend? The potential energy of this system is The gravitational potential energy, UgUg So what does this tell you about the elastic force?A conservative force!!! Hooke’s Law

25. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 25 Conservative and Non-conservative Forces When directly falls, the work done on the object by the gravitation force is The work done on an object by the gravitational force does not depend on the object’s path in the absence of a retardation force. How about if we lengthen the incline by a factor of 2, keeping the height the same?? Still the same amount of work Forces like gravitational and elastic forces are called the conservative force So the work done by the gravitational force on an object is independent of the path of the object’s movements. It only depends on the difference of the object’s initial and final position in the direction of the force. h l m  mgmg When sliding down the hill of length l, the work is N Total mechanical energy is conserved!! 1.If the work performed by the force does not depend on the path. 2.If the work performed on a closed path is 0.

26. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 26 Conservation of Mechanical Energy Total mechanical energy is the sum of kinetic and potential energies Let’s consider a brick of mass m at the height h from the ground The brick gains speed The lost potential energy is converted to kinetic energy!! What does this mean? The total mechanical energy of a system remains constant in any isolated systems of objects that interacts only through conservative forces: Principle of mechanical energy conservation m mgmg h What is the brick’s potential energy? What happens to the energy as the brick falls to the ground? m h1h1 By how much? So what? The brick’s kinetic energy increased And?

27. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 27 Example A ball of mass m at rest is dropped from the height h above the ground. a) Neglecting the air resistance, determine the speed of the ball when it is at the height y above the ground. b) Determine the speed of the ball at y if it had initial speed vi vi at the time of the release at the original height h.h. mgmg h m y m Using the principle of mechanical energy conservation Again using the principle of mechanical energy conservation but with non-zero initial kinetic energy!!! This result look very similar to a kinematic expression, doesn’t it? Which one is it? PEKE mgh mgy 0 0 mv 2 /2 mv i 2 /2 mv f 2 /2

28. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 28 Power Rate at which the work is done or the energy is transferred –What is the difference for the same car with two different engines (4 cylinder and 8 cylinder) climbing the same hill? – – The time… 8 cylinder car climbs up the hill faster! Is the total amount of work done by the engines different? NO Then what is different? The rate at which the same amount of work performed is higher for 8 cylinders than 4. Average power Unit? What do power companies sell? Energy Scalar quantity

29. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 29 Energy Loss in Automobile Automobile uses only 13% of its fuel to propel the vehicle. Why? 13% used for balancing energy loss related to moving the vehicle, like air resistance and road friction to tire, etc Two frictional forces involved in moving vehicles Coefficient of Rolling Friction;  =0.016 16% in friction in mechanical parts 4% in operating other crucial parts such as oil and fuel pumps, etc Air Drag Total Resistance Total power to keep speed v=26.8m/s=60mi/h Power to overcome each component of resistance 67% in the engine: Incomplete burning Heat Sound

30. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 30 Human Metabolic Rates

31. Tuesday, June 24, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 31 Ex. The Power to Accelerate a Car A 1.10x10 3 kg car, starting from rest, accelerates for 5.00s. The magnitude of the acceleration is a=4.60m/s 2. Determine the average power generated by the net force that accelerates the vehicle. What is the force that accelerates the car? Since the acceleration is constant, we obtain From the kinematic formula Thus, the average speed is And, the average power is