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# Work energy and potential energy

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### Work energy and potential energy

1. 1. Work The VERTICAL component of the force DOES NOT cause the block to move the right. The energy imparted to the box is evident by its motion to the right. Therefore ONLY the HORIZONTAL COMPONENT of the force actually creates energy or WORK. When the FORCE and DISPLACEMENT are in the SAME DIRECTION you get a POSITIVE WORK VALUE. The ANGLE between the force and displacement is ZERO degrees. When the FORCE and DISPLACEMENT are in the OPPOSITE direction, yet still on the same axis, you get a NEGATIVE WORK VALUE. IT simply means that the force and displacement oppose each other. The ANGLE between the force and displacement in this case is 180 degrees. When the FORCE and DISPLACEMENT are PERPENDICULAR, you get NO WORK!!! The ANGLE between the force and displacement in this case is 90 degrees.
2. 2. Scalar Dot Product?    W  F  x  Fx cos A dot product is basically a CONSTRAINT on the formula. In this case it means that F and x MUST be parallel. To ensure that they are parallel we add the cosine on the end. FORCE Displacement    W F x Fx    0  ; cos0 1    W  Fx cos  
3. 3.    W  F  x  Fx cos FORCE Displacement    W F x Fx    180  ; cos180 1      W F x f   cos  
4. 4.    W  F  x  Fx cos FORCE Displacement    W  F  x  Fx   90 ; cos90 0 W 0 J cos    
5. 5. Work = The Scalar Dot Product between Force and Displacement. So that means if you apply a force on an object and it covers a displacement you have supplied ENERGY or done WORK on that object. cos  F  r  F r  Wdisplacement vector  r    
6. 6. Example W  F  r W  F r cos A box of mass m = 2.0 kg is moving over a frictional floor has a force whose magnitude is F = 25 N applied to it at an angle of 30 degrees. The box is observed to move 16 meters in the horizontal direction before falling off the table. a) How much work does F do before taking the plunge? W  F  r W F r W cos 25 16 cos30   W  346.4 Nm W  346.4 J 
7. 7. Example What if we had done this in UNIT VECTOR notation? 21.65ˆ 12.5 ˆ F  i  j W F r F r W     ( ) ( ) x x y y     (21.65 16) (12.5 0) W  346.4 Nm W  346.4 J
8. 8. Work-Energy Theorem dx  ( )    W m v dv 2 2 2 v o v      2 2 ) 2 2 | ) ( 2 ( | 2 2 o v v mv mv W v v m v W m dv m v dv dt W m o o     2 K mv W   K 2 Kinetic Energy 1 Kinetic energy is the ENERGY of MOTION.
9. 9. Example Suppose the woman in the figure above applies a 50 N force to a 25-kg box at an angle of 30 degrees above the horizontal. She manages to pull the box 5 meters. a) Calculate the WORK done by the woman on the box b) The speed of the box after 5 meters if the box started from rest. W Fx   cos (50)(5)cos30   W  W   KE  mv W  v  v 2 2 (25) 2 1 2 1 216.5 J 4.16 m/s
10. 10. Conservative Force  Conservative Forces ◦ A force where the work it does is independent of the path A conservative force depends only on the position of the particle, and is independent of its velocity or acceleration. x y z A F B
11. 11. Something is missing…. Consider a mass m that moves from position 1 ( y1) to position 2 m,(y2), moving with a constant velocity. How much work does gravity do on the body as it executes the motion? W F r Fr gravity    W mg y y gravity   W mg y gravity    W U gravity   cos ( ) cos180 1 2    W F r Fr gravity    W mg y y gravity   W mg y gravity    W U gravity   cos ( ) cos0 2 1    Suppose the mass was thrown UPWARD. How much work does gravity do on the body as it executes the motion? In both cases, the negative sign is supplied
12. 12. The bottom line.. The amount of Work gravity does on a body is PATH INDEPENDANT. Force fields that act this way are CONSERVATIVE FORCES FIELDS. If the above is true, the amount of work done on a body that moves around a CLOSED PATH in the field will always be ZERO FRICTION is a non conservative force. By NON-CONSERVATIVE we mean it DEPENDS on the PATH. If a body slides up, and then back down an incline the total work done by friction is NOT ZERO. When the direction of motion reverses, so does the force and friction will do NEGATIVE WORK in BOTH directions.
13. 13. Nonconservative Forces  Examples of Nonconservative forces ◦ Friction ◦ Air resistance ◦ Tension ◦ Normal force ◦ Propulsion force of things like rocket engine  Each of these forces depends on the path
14. 14. Potential Energy mg h W Fx F mg x h  cos  ;  0,cos0 1   W  mgh  PE    Since this man is lifting the package upward at a CONSTANT SPEED, the kinetic energy is NOT CHANGING. Therefore the work that he does goes into what is called the ENERGY OF POSITION or POTENTIAL ENERGY. All potential energy is considering to be energy that is STORED!
15. 15. Conservation of Energy A B C D
16. 16. In Figure A, a pendulum is released from rest at some height above the ground position. It has only potential energy. In Figure C, a pendulum is at the ground position and moving with a maximum velocity. It has only kinetic energy. In Figure B, a pendulum is still above the ground position, yet it is also moving. It has BOTH potential energy and kinetic energy. In Figure D, the pendulum has reached the same height above the ground position as A. It has only potential energy.
17. 17. Hooke’s Law  If a ‘spring’ is bent, stretched or compressed from its equilibrium position, then it will exert a restoring force proportional to the amount it is bent, stretched or compressed.  Fs = kx (Hooke’s Law)  Related to Hooke’s Law, the work required to bend a spring is also a function of k and x.  Ws = (½ k)x2 = amount of stored strain energy ◦ K is the stiffness of the spring. Bigger means Stiffer ◦ x is the amount the spring is bent from equilibrium
18. 18. Based on Hooke’s Law  A force is required to bend a spring  The more force applied the bigger the bend  The more bend, the more strain energy stored Force (N) Relationship between F and x slope of the line = k x (m) Area under curve is the amount of stored strain energy Stored strain energy = ½ k x2
19. 19. Hooke’s Law from a Graphical Point of View Suppose we had the following data: x(m) Force(N) 0 0 0.1 12 0.2 24 0.3 36 0.4 48 0.5 60 0.6 72 F  kx  F  Slope of a F vs. x graph k x k s s Force vs. Displacement y = 120x + 1E-14 R2 = 1 80 70 60 50 40 30 20 10 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Displacement(Meters) Force(Newtons) k =120 N/m
20. 20. ELASTIC POTENTIAL ENERGY Recall that the force of an elastic spring is F = ks. It is important to realize that the potential energy of a spring, while it looks similar, is a different formula. Ve (where ‘e’ denotes an elastic spring) has the distance “s” raised to a power (the result of an integration) or 2 1 V ks e  2 Notice that the potential function Ve always yields positive energy.
21. 21. Elastic potential energy 2 W  F ( x ) dx  ( kx ) dx   W  kx dx  k xdx     0 0 0 2 2 | 1 2 | ( ) W U kx x W k spring x x x x x x      Elastic “potential” energy is a fitting term as springs STORE energy when there are elongated or compressed.
22. 22. Conservation of Energy in Springs
23. 23. Example A slingshot consists of a light leather cup, containing a stone, that is pulled back against 2 rubber bands. It takes a force of 30 N to stretch the bands 1.0 cm (a) What is the potential energy stored in the bands when a 50.0 g stone is placed in the cup and pulled back 0.20 m from the equilibrium position? (b) With what speed does it leave the slingshot? a F kx k k ) 30 (0.01) 3000 N/m s 2 b U kx k s c E E U K U mv v            v s B A s 2 2 (0.050) 2 1 2 1 ) 0.5( )(.20) 2 ) 1 300 J 109.54 m/s