1. Work and Power

2. Objectives 1.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) 2.The student will understand that energy is conserved (PH.6a) 3.The student will investigate and understand that energy can be transferred and transformed to provide usable work. (PH.8a,b)

3. Work Work is the force necessary to move an object a distance.

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

5. Unit of Work Need 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.75 m. Find the work done by the teacher for one rep (up and down) Known: Distance (d) =.75 m weight lifted = mg = 245kg x 9.8m/s 2  F y = ma y F Lift - mg = 0 F Lift = mg = (245 kg)(9.8 m/s 2 ) = 2403 N W Lift = F Lift d = (2403 N)(.75 m) x 2 = 3605 J mg d.5F L

7. What is the unit of work? 1.Newton 2.Meter 3.Joule 4.Newton meter 5.3 and 4

8. How much work is necessary to lift 10 kg 5m in the air? 1.10 N 2.50 J 3.490 J 4.4900 J 10 kg 5 m

9. Power Need 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)

10. 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/s 2 )(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

11. How much power is required to lift 10 kg, 5 m in the air in 10 s? 10 kg 5 m 1.49 w 2.490 J 3.490 w 4.4900 w

12. 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 (.75m x 10) so Work (W) = (Force) x (distance) x (#repetitions) (Assume no work done bringing weight down) Work = (2405 N)(.75 M)(10) W = 18,038 Joules (J) Power = Work/time P = (18,038J)/(25 s) P = 721 W

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

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

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

16. Bottom Line You can always use the equation Work = (Force)(Distance)(cosӨ) W=FdcosӨ Even if the Force is parallel to the displacement Force Displacement Ө = 0 o Cos0 o = 1

17. Work Done by a Constant Force Need 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

18. Problem Solving Techniques 1.FBD: Sketch the system and show the force that is doing the work (as well as others that may be involved) 2.Choose an x-y coordinate system - direction of motion should be one 3.Determine the force that is doing the work 4.Find the angle between the force doing the work and the displacement 5.Find the work done: W=(Force)(distance)cos  d = 40m FpFp FpFp 

19. Example: Work done on a crate F p = 100 N  = 37 deg Determine the work done by the force acting on the crate W p = F p dcos  = (100 N)(40 m)(cos37) = 3200J d = 40m FpFp FpFp 

20. How do we solve for friction? If F p = 100 N and  =45 o, what is F f ?  F x = ma x Must resolve F p into x-direction F p cos45 o – F f = 0 F f = F p cos45 o F f = 71 N FpFp  =45 o FfFf x F p cos45 o

21. What happens if we add friction? If F f = 71 N, what is W f ? Note the force friction and displacement are in the opposite direction, so …they are 180 o apart W f = Fd = F f dcos180 o (Where cos180 o = -1) = (71 N)(40 m)(-1) = -2840J FpFp  FpFp  FfFf FfFf d = 40m

22. What is the net work done on the crate? W net = W p + W f = 3200 J - 2000 J = 1200 J