Applications of Work and Energy Equations

1. m m Applications of Work and Energy Equations James Prescott Joule AP Physics B Lecture Notes

2. PHYSICS 3.0 m Problem #1 A student holds her physics textbook, which has a mass of 1.50 kg from the second floor which is 3.0 m from the ground. She then drops it. • How much work is done on the book by the student in simply holding it? • How much work will have been done by the force of gravity during • the time the book falls 3.0 m? • A known mass held, then dropped; • determine a) Wst and b) Wgr • Given: m = 1.50 kg, and h = 3.0 m; • Find: Wstand Wgr 3) Work / gravity RMc

3. PHYSICS -3.0 m mg Problem #1 4-5) a) Net work on the book by student: Since the book does not move, there is no displacement and therefore, no work done. Work = Fd cos(q) Wst = F(0) cos(q) = 0.0 J b) Net work on the book by force of gravity: While the book is falling, the Net force is equal to the weight (mg). Work = Fd cos(q) Wgr = (1.50kg)(-9.81m/s2)(-3.0m)cos(0) = +44 J RMc

4. 90.0 N 40.0 Problem #2 If a person pushes on the lawn mower with a constant force of 90.0 N at an angle of 40.0 to the horizontal: How much work does he do in pushing it a horizontal distance of 7.50 m? • A constant force at angle • thru distance; determine Wh • Given: F = 90.0 N, and q = 40.0; • d = 7.50 m • Find: Wh 3) Work / horizontal RMc

5. 40.0 F cos(q) = Fh 90.0 N 7.50 m 40.0 Problem #2 4-5) Net work on the mower by person: The force vector and the displacement vector must be parallel for work to be done. Work = Fd cos(q) Fh = (90.0N)(cos 40.0) Fh = 68.94 N Wh = (68.94N)(7.50m) = +517 J RMc

6. y mg cos(q) q x N Ff mg sin(q) q mg 1.20 m Problem #3 A 0.75 kg block slides down a 20.0 inclined plane with a uniform velocity. The horizontal length of the incline is 1.20 m. a) How much work is done by the force of friction on the block as it slides the total length of the incline? b) What is the net work done on the block? • Box slides at constant velocity against friction; • determine Wf , and Wnet • Given: m = 0.75 kg, and • q = 20.0;L = 1.20 m • Find: Wf and Wnet 3) Work / incline-friction RMc

7. y N x mg cos(q) Ff mg sin(q) Problem #3 Only two forces do work because there are only two forces parallel to the motion. 4-5) a) Net forces parallel to the motion: The force of kinetic friction Ff and the weight component down the incline mg sin(q). Work done by the frictional force: Ff = mN W = mNd(-1) W = Ffd cos(q) =Ffd cos(180) =Ffd (-1) Ff = mg sin (q) from uniform motion N = mgcos(q) from diagram RMc

8. d q L from right triangle incline W = -tan(q)mgcos(q) Problem #3 N = mgcos(q) from diagram W = mNd(-1) Therefore: W = -tan(20.0)mgL = -(.3639)(0.75kg)(9.81m/s2)(1.20m) W = -3.21 J RMc

9. y x Ff mg sin(q) Problem #3 4-5) b) Net forces parallel to the motion: Since the force of kinetic friction Ff is equal to weight component down the incline mg sin(q), the net force is equal to zero. The Net work done: Wnet = Fnetd cos(q) Wnet = (0)d cos(q) Wnet=0.0 J RMc

10. y x mg cos(q) N Ff mg sin(q) mg Problem #3 A car of mass m is on an icy driveway inclined at an angle of 20.0. Determine the acceleration of the car (assume the incline is frictionless). • Unknown mass down frictionless incline; • determine acceleration. • Given:  = 20.0; • Find: am 3) Newton’s 2nd Law / inclined frictionless q RMc

11. Work Done by a Constant Force Work = Fd cos(q) RMc

12. Fcos(q) Work Equation F q d Work = Fd cos(q) RMc

13. Units of Work SystemUnit of Work SI newton-meter Joule (J) cgs dyne-centimeter erg British foot-pound RMc

14. Work-Kinetic Energy Theorem vf vi F m x The net work done on an object by a net force acting on it is equal to the change in kinetic energy of the object RMc

15. Kinetic Energy F m x Work-Kinetic Energy Theroem A net force acts on a moving object vi vf RMc

16. Work Done Against Friction Friction acts on moving object m m d Ff RMc

17. The Force to Stretch a Spring Hooke’s Law RMc

18. x F F kx 0 x Work RMc

19. m Work Done by Gravity Object falls in a gravitational field mg h m Wg = mgh RMc

20. Work and Energy Equations Work done by force (F) = Kinetic energy = Work done by gravity = Work done by a spring = Work done against friction = RMc

21. Gravity and Kinetic Energy Ball drooped from rest Find the final velocity of the ball: h v RMc

22. m Spring and Kinetic Energy Find the final velocity of the block: Work done by a spring v m k x RMc

23. m Spring and Friction Find the displacement of the block: Work done by friction v = 0 m k x d = ? RMc

24. vo = 0 m m d h q V = ? h = d sin(q) Kinetic Energy and the Incline Plane Find the final velocity of the block: RMc

25. m m Gravitational Potential Energy Work Done due to Gravity h RMc

26. k Elastic Potential Energy k x RMc

27. Conservation of Mechanical Energy The mechanical energy (Emec) of a system is When no external forces act on a system Conservation of Mechanical Energy RMc

28. Potential and Kinetic Energy Conservation of Mechanical Energy U K U K U K U K K U

29. Power Energy can be transferred from one form to another through the process of work. The rate at which work is done is called power Average power RMc

30. Work and Energy END