1. Work and Energy

2. Work Work means many things in everyday life. However, in Physics, work is defined one way. Work = the product of the magnitude of the displacement times the component of the force parallel to the displacement. W =, where is the force parallel to the displacement, d. (Remember the boat from our last test)

3. The power of θ The force parallel to the displacement is the x component of the force and is given by Fcosθ So, W = Fdcosθ When θ = 0, cosθ = 1 and all of the force goes into work and W =Fd

4. Units of Work Force x distance is going to give us Newtons x meters (Nm) This unit takes the name joule (J): 1J = 1Nm – Named in honor of James Prescott Joule an English brew master, who discovered the link between mechanical energy and heat. (more on that when we get to thermodynamics)

5. Other units of work Any units of force and distance will give you a unit for work. One such unit is the erg. 1erg = 1dyne x cm In British units (which Americans use as well) the unit is the foot-pound. 1J = 10 7 erg = 0.7376ft*lb

6. Unemployed Will a force always create work? If not, under what circumstances will there be no work done?

7. Example A 50kg crate is pulled 40m along a horizontal floor by a constant force exerted by a person, F P = 100N, which acts at a 37 o angle. The floor is rough and exerts a friction force F fr = 50N. Determine the work done by each force acting on the crate, and the net work done on the crate.

8. Solution We have 4 forces: push, friction, weight, and normal. Weight and normal have a θ of 90 (weight is actually -90 or 270). cos90 = 0, so Fdcosθ = 0. The weight and normal forces do NO WORK. W push = F push dcosθ = 100N(40m)cos37 = 3200J W fr = F fr d cos180 = (50N)(40m)(-1) = -2000J

9. Solution part 2 W net = W g + W N + W P + W fr = 0 + 0 + 3200J -2000J = 1200J OR W net = F net,x d = (F P cosθ – F fr )d = (100N cos37 – 50N)(40m) = 1200J

10. Kinetic Energy Is the energy associated with an object in motion KE = 1/2mv 2

11. Kinetic Energy and Work Remember, W= Fd F = ma So W = mad d= ∆x So W = ma∆x

12. The Math Remember way back into chapter 2, v f 2 = v i 2 + 2a∆x Solve for a∆x, plug in to get, W = 1/2mv f 2 – 1/2mv i 2

13. Kinetic Energy and Work Cont. …Therefore, W = ∆KE

14. Example A 145g baseball is thrown with a speed of 25m/s. What is its kinetic energy? How much work was done on the ball to make it reach this speed, if it started from rest?

15. Solution KE = 1/2mv 2 = ½(0.145kg)(25m/s) 2 = 45J Because the ball started at rest it had 0J of kinetic energy at the beginning, so the net work done is equal to the kinetic energy, 45J.

16. Example How much work is required to accelerate a 1000kg car from 20m/s to 30m/s?

17. Solution W = KE 2 – KE 1 = 1/2mv 2 2 – 1/2mv 1 2 = ½(1000kg)(30m/s) 2 – ½(1000kg)(20m/s) 2 = 2.5 x 10 5 J

18. Potential Energy Is associated with an object that has the potential to move because of its location relative to some other location.

19. Gravitational Potential Energy PE g = mgh, where m is the mass, g is gravity, and h is the height of the object. But, the height above what?

20. Example A 0.1g pencil slowly rolls off of a 1m high desk and onto the floor. What was the pencil’s potential energy when it was: A. on the desk w.r.t the desk? B. on the desk w.r.t the floor? C. on the floor w.r.t the desk?

21. Solution First, we need kg. Why? – 0.1g * (1kg/1000g) = 1.0 x 10 -4 kg A. PE = mgh = 1.0 x 10 -4 kg(9.8m/s 2 )(0m) = 0J B. PE = mgh = 1.0 x 10 -4 kg(9.8m/s 2 )(1m) = 9.8 x 10 -4 J C. PE = mgh =1.0 x 10 -4 kg(9.8m/s 2 )(-1m) = -9.8 x 10 -4 J Can PE be negative?

22. Elastic Potential Energy The potential energy stored in a compressed or stretched object, like a spring. The length of a spring when no forces are acting on it is called the relaxed length.

23. More on Springs The spring constant, k, refers to how “stiff” the spring is. The higher the k the harder it is to squeeze the spring. If a spring is hard to squeeze, then there is a lot of energy in the spring when compressed. The distance the spring is squeezed or stretched from its relaxed length we call x. You need more energy to increase x.

24. The Point PE elastic = ½kx 2

25. Example A spring has a spring constant of 200N/m and is compressed 10cm from its relaxed length. What potential energy is stored in the spring?

26. Solution PE elastic = ½kx 2 PE = ½(200N/m)(0.1m) 2 = 1J Unit Check: N/m *m 2 = Nm = J

27. Roller coasters Roller coasters are powered by gravitational potential energy. At the beginning of the ride, work is done on the coaster to give it potential energy. Then work is done by the coaster to turn its potential energy into kinetic energy (it falls down)

28. More on coasters A coaster that falls to the bottom of the hill has turned all of its potential energy into kinetic. If the coaster goes back up another hill, it loses kinetic energy (slows down) and turns it back into potential energy (gets higher).

29. Ponder this Can a coaster climb a hill higher than the one it started at? Explain. What happens to the energy of the coaster when it is halfway up a hill? Halfway down?

30. Mechanical Energy As you should realize by now, potential and kinetic energy are closely related. KE and PE combine to make a new quantity, E, the total mechanical energy. E = KE + PE

31. Conservation of energy Mechanical energy is conserved = it is not created or destroyed. ΔKE + ΔPE = 0 KE 2 + PE 2 = KE 1 + PE 1 E 2 = E 1

32. Example: Falling rock If the original height of a stone is 3.0m, calculate the stone’s speed when it has fallen to 1.0m above the ground.

33. solution v 1 = 0m/s h 1 = 3.0m h 2 = 1.0m g = 9.8m/s 2 1/2mv 1 2 + mgh 1 = 1/2mv 2 2 + mgh 2 0 + m(9.8m/s 2 )(3.0m) = ½(m)(v 2 2 ) + m(9.8m/s 2 )(1.0m) We can cancel all of the m’s and solve for v 2 2 v 2 2 = 2[(9.8m/s 2 )(3.0m) – (9.8m/s 2 )(1.0m)]=39.2m 2 /s 2 v 2 = 6.3 m/s

34. Power In everyday life, when we say that is powerful we usually mean that it can put out a lot of force. In physics, the official definition of power is this: the rate at which work is done

35. The Units of Power Work/time would give us J/s A J/s has a short name, the Watt, in honor of James Watt, the inventor of the steam engine. 1J/s = 1W Yes, this is the Watts like in light bulbs.

36. Horse Power Like all American units this one has a really dumb origin story. James Watt needed a way to explain how powerful his new steam engine was. He ran an experiment and concluded that a “good horse” can work all day at a rate of 360ft*lb/s. He multiplied this by 1.5 just to make the horse look better and got 550ft*lb/s

37. Horse Power So, Watt declared 1 horse power (1hp) to be equal to 550ft*lb Putting this into SI units gives us: 1hp = 746W

38. Example A 70kg jogger runs up a long flight of stairs in 4.0s. The vertical height of the stairs is 4.5m. A.) Find the jogger’s power output in both Watts and horsepower. B.) How much energy did this take?

39. Solution The work done is against gravity, so W = mgy. P = W/t = mgy/t = (70kg)(9.8m/s 2 )(4.5m) / 4.0s = 770W Because 1 hp = 746W the person’s power out put is just over 1hp. The energy required is E = Pt (because work is energy transfer). E = (770J/s)(4.0s) = 3100J

40. Firing Mah Lazer The Nova laser at Lawrence Livermore National Lab has 10 beams, each beam has a power output greater than all of the power plants in the United States. Where does this power come from?

41. Power in terms of velocity It is often useful to write power in terms of v instead of work. Here’s how: P = W/t = Fd/t (d/t = v) so P = F Note: this is the average velocity, d/t

42. Example Calculate the power required of a 1400kg car under the following circumstances: A. the car climbs a 10 degree hill at 80km/hr B. the car accelerates along a flat road from 90km/hr to 110km/hr in 6.0s. Assume F fr = 700N