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- 1. Presented by: Karen A. Adelan BSE 3 Classical Mechanics
- 2. Why Energy?  Why do we need a concept of energy?  The energy approach to describing motion is particularly useful when Newton’s Laws are difficult or impossible to use  Energy is a scalar quantity. It does not have a direction associated with it
- 3. What is Energy?  Energy is a property of the state of a system, not a property of individual objects: we have to broaden our view.  Some forms of energy:  Mechanical:  Kinetic energy (associated with motion, within system)  Potential energy (associated with position, within system)    Chemical Electromagnetic Nuclear  Energy is conserved. It can be transferred from one object to another or change in form, but cannot be created or destroyed
- 4. Kinetic Energy  Kinetic Energy is energy associated with the state of motion of an object  For an object moving with a speed of v KE 1 2 mv 2  SI unit: joule (J) 1 joule = 1 J = 1 kg m2/s2
- 5. Work W 1 2 1 2 Fx x  Start with 2 mv 2 mv0 Work “W”  Work provides a link between force and energy  Work done on an object is transferred to/from it  If W > 0, energy added: “transferred to the object”  If W < 0, energy taken away: “transferred from the object”
- 6. Work Unit  This gives no information about the time it took for the displacement to occur  the velocity or acceleration of the object  Work is a scalar quantity 1 2 1 2  SI Unit mv mv0 ( F cos 2 2  Newton • meter = Joule N • m = J  J = kg • m2 / s2 = ( kg • m / s2 ) • m  W ( F cos ) x ) x
- 7. Work: + or -?  Work can be positive, negative, or zero. The sign of the work depends on the direction of the force relative to the displacement W      ( F cos ) x Work positive: W > 0 if 90°> > 0° Work negative: W < 0 if 180°> > 90° Work zero: W = 0 if = 90° Work maximum if = 0° Work minimum if = 180°
- 8. Example: When Work is Zero  A man carries a bucket of water horizontally at constant velocity.  The force does no work on the bucket  Displacement is horizontal  Force is vertical  cos 90° = 0 W ( F cos ) x
- 9. Example: Work Can Be Positive or Negative  Work is positive when lifting the box  Work would be negative if lowering the box  The force would still be upward, but the displacement would be downward
- 10.  F Work Done by a Constant Force  The work W done on a system by an agent exerting a constant force on the system is the product of the magnitude F of the force, the magnitude Δr of the displacement of the point of application of the force, and cosθ, where θ is the angle between the force and displacement vectors: W  F  r  F  r  r II I  0F WI WII F r  F  r  r F r cos III WIII F r IV WIV F r cos February 11, 2014
- 11. Work Done by Multiple Forces  If more than one force acts on an object, then the total work is equal to the algebraic sum of the work done by the individual forces Wnetnet W  Wby by individual forces Windividual forces Remember work is a scalar, so this is the algebraic sum Wnet Wg WN WF Wnet Wg WN WF ( F cos ) r ( F cos ) r February 11, 2014
- 12. Work and Multiple Forces  Suppose µk = 0.200, How much work done on the sled by friction, and the net work if θ = 30° and he pulls the sled 5.0 m ? ( f k cos180 ) x W fric k N x k fk x (mg F sin ) x (0.200)(50.0kg 9.8m / s 2 1.2 102 N sin 30 )(5.0m) 4.3 102 J Wnet WF W fric WN Wg 5.2 102 J 90.0 J 4.3 102 J 0 0
- 13. Kinetic Energy object is the energy which it possesses due to its motion.[1] It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest. In classical mechanics, the kinetic energy of a non-rotating object of mass m traveling at a speed v is ½ mv². In relativistic mechanics, this is only a good approximation when v is much less than the speed of light.
- 14. Work-Kinetic Energy Theorem Is the net work done on an object is equal to the change in the kinetic energy of the object. Wnet = ∆KE Net work is equal to kinetic energy
- 15. • Kinetic energy depends on speed and mass: KE = ½mv2 Kinetic energy = ½ x mass x (speed)2 KE is a scalar quantity, SI unit (Joule)
- 16. • TRY TO SOLVE: Ex. A 7.00 kg bowling ball moves at 3.00 m/s. how much kinetic energy does the bowling ball have? how fast must a 2.24 g table-tennis ball move in order to have the same kinetic energy as the bowling ball? Is the speed reasonable for a table-tennis ball? Given: the subscript b and t indicate the bowling ball and the table-tennis ball, respectively. Mb = 7.00 kg Mt = 2.24g Vb = 3.00m/s Unknown: KEb = ? Vt = ?
- 17. Work-Kinetic Energy Theorem  When work is done by a net force on an object and the only change in the object is its speed, the work done is equal to the change in the object’s kinetic energy   Speed will increase if work is positive Speed will decrease if work is negative Wnet 1 mv 2 2 1 2 mv0 2
- 18. Work-Energy Theorem W= KE W=KEf-KEi “In the case in which work is done on a system and the only change in the system is in its speed, the work done by the net force equals the change in kinetic energy of the system.”
- 19. • The Work-Kinetic Energy Theorem can be applied to non isolated systems • A non isolated system is one that is influenced by its environment (external forces act on the system)
- 20. Work and Kinetic Energy  The driver of a 1.00 103 kg car traveling on the interstate at 35.0 m/s slam on his brakes to avoid hitting a second vehicle in front of him, which had come to rest because of congestion ahead. After the breaks are applied, a constant friction force of 8.00 103 N acts on the car. Ignore air resistance. (a) At what minimum distance should the brakes be applied to avoid a collision with the other vehicle? (b) If the distance between the vehicles is initially only 30.0 m, at what speed would the collisions occur?
- 21. Work and Kinetic Energy 3 v 0 35/,s m s m 1. m. 10 10 kg3 f 8 f k8 10 310 10 3 v 8 N v0 0 35 .00.mms.0, v /0,,0, m 0,00001.00 3 ,10k, kg, .00 .00.00 3 N N kg f k 35 / v 1  (a) We know v  Find the minimum necessary stopping distance 1 2 1 2 Wnet W fric Wg WN W fric mv mv 1 22 f 2 i f k x 0 1 mv0 2 f k x 0 2 mv0 2 1 1 103 kg)(35.0m / s) 2 (8.00 108.N ) 103 N ) x (1.001.00 103 kg)(35.0m / s) 2 ( 00 x ( 2 2 3 x 76.6.6m x 76m February 11, 2014
- 22. Work and Kinetic Energy    x 30 .0m, v0 35 .0m / s, m 1.00 10 3 kg, f k (b) We know Find the speed at impact. Write down the work-energy theorem: 1 2 Wnet W fric fk x mv f 2 2 2 v 2 v0 fk x f m v2 f vf 1 2 mvi 2 2 (35 m / s) 2 ( )(8.00 10 3 N )(30 m) 1.00 10 3 kg 27.3m / s 8.00 10 3 N 745 m 2 / s 2

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