Work and Energy

1. Work and Energy

2. Definitions • Work – the product of force and the component of displacement in the direction of the force. - work is a scalar quantity - Without motion there is no work. W = F ∙ d The unit of work is the newton meter, which is called a joule (J) (in honor of English Scientist James Prescott Joule)

3. Example • How much work is done on an object if a force of 30 Newtons [south] displaces the object 200 meters [south]? • Solution: W = F∙d = (30 N[S])(200 m [S]) = 6000 J

4. Work (cont’d) • Suppose force and displacement are not in the same direction. • Work is defined to be the product of the force in the direction of the displacement and the displacement F W = (Fcosθ)∙d d θ object

5. example • As Alex pulls his red wagon down the sidewalk, the handle of the wagon makes an angle of 60 degrees with the pavement. If Alex exerts a force of 100 Newtons along the direction of the handle, how much work is done when the displacement of the wagon is 20 meters along the ground?

6. Power

7. Definition • Power – the rate at which work is done Power is also a scalar quantity, and its unit is Joules per second (J/s), also known as a watt(W).

8. Example • If 300 J of work is performed on an object in 1.0 minute, what is the power expended on the object? P = W / t P = 300J / 60 sec P = 5 W

9. Example 2 • A 200 N force is applied to an object that moves in the direction of the force. If the object travels with a constant velocity of 10m/s, calculate the power expended on the object. P = 200N (10 m/s) P = 2000 W

10. Hw • Read pg. 80-83 Do # 1-25

11. Energy Mechanical Energy

12. Kinetic Energy • Work done on an object changes its Kinetic Energy. • Therefore, W = KEf – KEi = ΔKE. • The formula for Kinetic Energy is KE = ½ mv2

13. Example • A 10 kg object subjected to a 20. N force moves across a horizontal, frictionless surface in the direction of the force. Before the force was applied, the speed of the object was 2.0m/s. When the force is removed the object is traveling at 6.0 m/s. Calculate the following quantities: (a) KEi , (b) KEf , (c)ΔKE, (d) W, and (e) d.

14. Solution (a) KEi = ½ mvi2 = ½ (10. kg)(2.0 m/s)2 = 20. J (b) KEf = ½ mvf2 = ½ (10. kg)(6.0 m/s)2 = 180 J

15. Solution (cont’d) (c) ΔKE = KEf – KEi = 180 J – 20. J = 160 J (e) W = F∙d d = W/F = 160 J / 20. N = 8.0 m (d) W = ΔKE = 160 J

16. Gravitational Potential Energy • An object decreases its gravitational Potential Energy (PE) as it moves closer to the Earth • To calculate the change in PE of an object we measure the work done on the object. The force needed to overcome gravity is Fg = mg. Therefore, since W = Fg ∙d , PE is defined as ΔPE = mg Δ h where Δh represents change in vertical displacement above the earth.

17. Example • A 2.00 kg mass is lifted to a height of 10.0 m above the surface of the Earth. Calculate the change in the PE of the object. • Solution ΔPE = mg Δ h = (2.00 kg) (9.8 m/s2)(10.0m) = 196 J

18. NOTE • For a change in gravitational energy to occur, there must be a change in the vertical displacement of an object; if it is moved only horizontally, the ΔPE = 0. • If an object is moved up an inclined plane, its potential energy change is measured by calculating only its verticaldisplacement; the horizontal part does not change its PE

19. Home work • Review book : read pg 84 Do # 26-38 (PE) • Read pg 92-93 Do #57 -66 (KE)

20. Conservation of Mechanical Energy

21. Conservation of Mechanical Energy • In a system, the sum of PE and KE (the total mechanical energy) is constant (i.e. conserved); a change in one is accompanied by an opposite change in the other. ΔPE = -ΔKE PEi + KEi = PEf +KEf

22. example • A 0.50 kg ball is projected vertically and rises to a height of 2.0 meters above the ground. Calculate: (a) the increase in the ball’s PE (b) the decrease in the ball’s KE (c) the initial KE (d) the initial speed of the ball

23. solution • The increase in the ball’s PE ΔPE = mg Δ h = (0.50 kg)(9.8 m/s2)(2.0 m) = 9.8 J (b) The decrease in the ball’s KE ΔKE = -ΔPE = -9.8 J

24. Solution cont’d (c) The initial KE • Recall that at its highest point, the speed of the ball is zero; therefore its KE is zero. So the initial KE represents the change in KE of the object . ΔKE = KEf – KEi -9.8J = 0 - KEi KEi = 9.8 J

25. Sol’n cont’d • The initial speed of the ball KEi = ½ mvi2 Solving for vi = sqrt (2KEi /m) = sqrt [ 2(9.8J)/(0.50kg) ] = 6.3 m/s

26. Pendulum Observe that the falling motion of the bob is accompanied by an increase in speed. As the bob loses height and PE, it gains speed and KE; yet the total of the two forms of mechanical energy is conserved

27. example • A pendulum whose bob weighs 12 N is lifted a vertical height of 0.40 m from its equilibrium position. Calculate: (a) change in PE between max height and equilibrium height (b) Gain in KE and (c) the velocity at the equilibrium point.

28. solution • Take PE at lowest point to be zero ΔPE = mg Δ h = FgΔ h = (12 N)(-0.40m) = -4.8 J (b) ΔKE = -ΔPE = -(-4.8 J) = 4.8 J

29. Sol’n cont’d ( c ) First must calculate mass of bob Fg = mg m = Fg / g = 12N / 9.8m/s2 = 1.2 kg Then calculate velocity KE= ½ mv2 v = sqrt (2KE/m) = sqrt [ 2(4.8J)/(1.2kg) ] = 2.8 m/s

30. Elastic Potential Energy and Springs

31. Hooke’s Law • The English scientist Robert Hooke was able to show that the magnitude of a force (F) is directly proportional to the elongation (stretch) of compression of a spring (x) within certain limits. Fs = kx k – spring constant – unit is the newton per meter (N/m) Note: the greater the constant, the stiffer the spring

32. Elastic Potential Energy Spring is attached to a wall. If a force is applied and stretch the string to the right, work has been done. This work has been converted into the spring’s potential energy – Elastic Potential Energy PEs = ½ kx2

33. Hooke’s law and Potential Energy Area under the graph equals the work done. Fs = kx W

34. Example • A spring whose constant is 2.0 N/m is stretched 0.40 m from its equilibrium position. What is the increase in the elastic potential energy of the spring? Solution PEs = ½ kx2 = ½ (2.0 N/m ) (0.40 m)2 = 0.16 J

35. Elastic and Inelastic Collisions • In an elastic collision BOTH kinetic energy and momentum are conserved p1i + p2i = p1f + p2f KE1i+ KE2i = KE1f +KE2f p1i p2i p1f p2f 1 2 1 2 1 2 KE1i KE2i KE2f KE1f

36. Inelastic collisions • In inelastic collisions, the kinetic energy that is “lost” is converted into internal energy (Q) of the objects by frictional forces. • These systems are called nonideal mechanical systems. • The energy is constant ET = PE + KE + Q A change in the internal energy of an object is usually accompanied by a change in temperature

37. Simple Machines and Work

38. Simple Machines • A simple machine is a device that allows work to be done and offers and advantage to the user. Ex: pulleys, levers, inclined planes, wheels and axles and screwdrivers Win = Wout Fin ∙ din = Fout ∙ dout Fout / Fin = din / dout Mechanical Advantage (MA): Fout / Fin Ideal MA (IMA) assumes no friction (use din / dout ) Actual MA (AMA) is always less than IMA (use Fout / Fin ) The efficiency of machine is AMA/IMA ratio and this value is always less than 100%

39. example • A 100 N object is moved 2 m up an inclined plane whose end is lifted 0.5 m from the floor. If a force of 50 N is needed to accomplish this task, calculate the (a) IMA (b) AMA And (c) efficiency of the inclined plane

40. solution Input force (Fin ) = 50 N (force needed to move object) Output force (Fout ) = 100N (force that has been lifted) Input distance (din ) = 2m (distance moved along plane) Output distance (dout ) = 0.5m (distance object is raised) • IMA = din / dout = 2m / 0.5m = 4 • AMA = Fout / Fin = 100N / 50 N = 2 • efficiency = AMA/IMA = 2/4 = 0.5 (50%)

41. Internal Energy and Work

42. Internal Energy (Q) • Internal Energy of a system is the total kinetic and potential energies of the atoms and molecules that make up the system. • Recall: a change in the internal energy of an object is usually accompanied by a change in its temperature.

43. example • Force is used to move an object along a horizontal table at constant speed. • Has work been done? • Is there a change in Kinetic Energy? Why or why not? • Is there a change in Potential Energy? Why or why not? • How was work used? Yes W = F ∙ d No – speed is constant No – table is horizontal - so no change in height Used to overcome friction between object and table so internal energy of the object-table system has been increased by work done.

44. The Laws of Thermodynamics The study of the relationships among heat, work, and energy in the universe

45. The first law of Thermodynamics • Energy can neither be created nor destroyed. It can only change forms. • The change in the internal energy of a system (ΔU) is equal to the heat (Q) that the system absorbs (or releases) minus the work (W) it does (or has done on it) ΔU = Q – W

46. 2nd Law of Thermodynamics • Result if the work of the French physicist Nicolas Carnot with heat engines. • Law states heat cannot flow from colder object to a warmer one without work being done on the system. • Ex. Refrigerators must be run by motors in order to withdraw heat from objects placed in them • No heat engine can be 100% efficient. Some of the heat absorbed by the engine must be lost in the random motion of its molecules. • Entropy is the measure of this disorder

47. 3rd law of Thermodynamics • As temperature approaches absolute zero, (0 K) the entropy of a system approaches a constant minimum • The efficiency of a heat engine depends on its operating temperatures; engine would reach 100 % efficiency only at 0 K. • Engine cannot be completely efficient, therefore 0K cannot be reached.