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1. Work and Energy

2. Objectives • The student will investigate and understand the interrelationships among mass, distance, force and time through mathematical and experimental process. Key concepts include work, power and energy (PH.5 g) • The student will understand that energy is conserved (PH.6a) • The student will investigate and understand that energy can be transferred and transformed to provide usable work. (PH.8a,b)

3. Work Work Work is the force necessary to move an object a distance. Not Work

4. WorkNeed to know! Work = Force x distance moved (If the force is in the same direction as the displacement)

5. Unit of WorkNeed to Know Work is measured in newton-meters. A special name is given to this unit: Joule (J) = 1 Newton-meter Work is a scalar NOT a vector

6. “Working” out A man benches 585 lb (245 kg) The distance from his chest to the Top of the lift is 0.75 m. Find the work done by the teacher for one rep (up and down) Known: Distance (d) = 0.75 m weight lifted = mg = 245kg x 9.8m/s2 Fy = may FLift - mg = 0 FLift = mg = (245 kg)(9.8 m/s2) = 2403 N WLift = FLiftd = (2403 N)(0.75 m) = 1802.5 J (continued) mg d .5FL .5FL

7. “Working” out (cont.) FLower - mg = 0 FLower = mg = (245 kg)(9.8 m/s2) = 2403 N WLower = FLowerd = (2403 N)(-0.75 m) = -1802.5 J Total Work = WLift + WLower Total Work = 1802.5 J + (-1802.5 J) Total Work = 0 J mg d .5FL .5FL

8. What is the unit of work? • Newton • Meter • Joule • Newton meter • 3 and 4

9. 10 kg 5 m How much work is necessary to lift 10 kg 5m in the air? • 10 N • 50 J • 490 J • 4900 J

10. Recall • If we applied a force to an object • Work = Force x Distance • Previously in our examples, the force aligned with the distance Force Force Distance

11. BUT, what happens if the force and the distance are …. NOT in the Same Direction Force Force Distance

12. If this is the case • We must use the component of the force in the direction of the displacement Work = Force x Distance x cosӨ Need to Know F F Ө FcosӨ displacement

13. Bottom Line • You can always use the equation Work = (Force)(Distance)(cosӨ) W=FdcosӨ • Even if the Force is parallel to the displacement Force Cos0o = 1 Ө = 0o Displacement

14. Work Done by a Constant ForceNeed to Know • Work: the product of the magnitude of the displacement times the component of the force parallel to the displacement W = Fdcos Where F is the magnitude of the constant force, d is the magnitude of the displacement of the object, and  is the angle between the directions of the force and the displacement

15. Graphical Analysis of WorkNeed to Know • We can generate a graph of force versus distance. For a constant force, Fapplied F, W=Fxd N Area=LxW Distance, m x We can determine amount of work by calculating the area of the force-distance rectangle.

16. Graphical Analysis of WorkNeed to Know • Force is not always constant, but we can determine the work by calculating the area under the Force-distance curve: F, W= (1/2)Fxd N Area=(1/2) base x height Distance, m

17. Fp Fp  d = 40m Problem Solving Techniques • FBD: Sketch the system and show the force that is doing the work (as well as others that may be involved) • Choose an x-y coordinate system - direction of motion should be one • Determine the force that is doing the work • Find the angle between the force doing the work and the displacement • Find the work done: W=(Force)(distance)cos

18. Fp Fp  d = 40m Example: Work done on a crate Fp = 100 N  = 37 deg Determine the work done by the force acting on the crate Wp = Fpdcos = (100 N)(40 m)(cos37) = 3200J

19. PowerNeed to know • Power is the rate at which work is done P = average power = work/time = W/t Unit: Watt(w) = Joule/sec (J/s)

20. Power Example: Running Stairs A 70 kg student runs up a flight of stairs in 4.0s. The vertical height of the stairs is 4.5 m. Estimate the student’s power output in watts Know: mass = 70 kg; time = 4.0 s; y = 4.5 m The work is done against gravity: W= Force x distance; where force = mg And distance equals vertical distance y Work = (mass) x (gravity) x (y) W = (70 kg)(9.8m/s2)(4.5m) W = 3087 Joules (J) P = Work/time P = 3087 J/ 4.0 s P = 772 W (Recall 746 W = 1hp) = 1.03 hp mg y = 4.5

21. 10 kg 5 m How much power is required to lift 10 kg, 5 m in the air in 10 s? • 49 w • 490 J • 490 w • 4900 w

22. Power Example: Bench Press If a teacher benches 245 kg (weight = mg = 2405N) 0.75 m ten times in 25 seconds, estimate the power in his chest and arms The teacher moves the weight (Force required = mg = 2405N) a total distance of 7.5 m (0.75m x 10) so Work (W) = (Force) x (distance) x (#repetitions) (Assume no work done bringing weight down) Work = (2405 N)(0.75 M)(10) W = 18,038 Joules (J) Power = Work/time P = (18,038J)/(25 s) P = 721 W